You are given a sequence of $n$ digits $d_1d_2 \dots d_{n}$. You need to paint all the digits in two colors so that:

- each digit is painted either in the color $1$ or in the color $2$;
- if you write in a row from left to right all the digits painted in the color $1$, and then after them all the digits painted in the color $2$, then the resulting sequence of $n$ digits will be non-decreasing (that is, each next digit will be greater than or equal to the previous digit).

For example, for the sequence $d=914$ the only valid coloring is $211$ (paint in the color $1$ two last digits, paint in the color $2$ the first digit). But $122$ is not a valid coloring ($9$ concatenated with $14$ is not a non-decreasing sequence).

It is allowed that either of the two colors is not used at all. Digits painted in the same color are not required to have consecutive positions.

Find any of the valid ways to paint the given sequence of digits or determine that it is impossible to do.

**Input**

The first line contains a single integer $t$ ($1 \le t \le 10000$) — the number of test cases in the input.

The first line of each test case contains an integer $n$ ($1 \le n \le 2\cdot10^5$) — the length of a given sequence of digits.

The next line contains a sequence of $n$ digits $d_1d_2 \dots d_{n}$ ($0 \le d_i \le 9$). The digits are written in a row without spaces or any other separators. The sequence can start with 0.

It is guaranteed that the sum of the values of $n$ for all test cases in the input does not exceed $2\cdot10^5$.

**Output**

Print $t$ lines — the answers to each of the test cases in the input.

If there is a solution for a test case, the corresponding output line should contain any of the valid colorings written as a string of $n$ digits $t_1t_2 \dots t_n$ ($1 \le t_i \le 2$), where $t_i$ is the color the $i$-th digit is painted in. If there are several feasible solutions, print any of them.

If there is no solution, then the corresponding output line should contain a single character ‘-‘ (the minus sign).

**Example**

input

5 12 040425524644 1 0 9 123456789 2 98 3 987

output

121212211211 1 222222222 21 -

Note

In the first test case, $d=040425524644$. The output $t=121212211211$ is correct because $0022444$ (painted in $1$) concatenated with $44556$ (painted in $2$) is $002244444556$ which is a sorted sequence of $n$ given digits.

**Solution:**

#include <bits/stdc++.h> using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(0); int tt; cin >> tt; while (tt--) { int n; cin >> n; string s; cin >> s; vector<int> a(n); for (int i = 0; i < n; i++) { a[i] = (int) (s[i] - '0'); } bool found = false; for (int split = 0; split <= 9; split++) { vector<int> color(n, 0); int max1 = -1; for (int i = 0; i < n; i++) { if (a[i] < split) { color[i] = 1; max1 = max(max1, i); } if (a[i] > split) { color[i] = 2; } } for (int i = 0; i < n; i++) { if (a[i] == split) { if (i > max1) { color[i] = 1; } else { color[i] = 2; } } } int last = -1; bool ok = true; for (int r = 1; r <= 2; r++) { for (int i = 0; i < n; i++) { if (color[i] == r) { if (a[i] < last) { ok = false; } last = a[i]; } } } if (ok) { for (int i = 0; i < n; i++) { cout << color[i]; } cout << '\n'; found = true; break; } } if (!found) { cout << "-" << '\n'; } } return 0; }