Bear Limak has n colored balls, arranged in one long row. Balls are numbered 1 through n, from left to right. There are n possible colors, also numbered 1 through n. The i-th ball has color t i.
For a fixed interval (set of consecutive elements) of balls we can define a dominant color. It’s a color occurring the biggest number of times in the interval. In case of a tie between some colors, the one with the smallest number (index) is chosen as dominant.
There are non-empty intervals in total. For each color, your task is to count the number of intervals in which this color is dominant.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 5000) — the number of balls.
The second line contains n integers t 1, t 2, …, t n (1 ≤ t i ≤ n) where t i is the color of the i-th ball.
Output
Print n integers. The i-th of them should be equal to the number of intervals where i is a dominant color.
Examples
input
4
1 2 1 2
output
7 3 0 0
input
3
1 1 1
output
6 0 0
Note
In the first sample, color 2 is dominant in three intervals:
- An interval [2, 2] contains one ball. This ball’s color is 2 so it’s clearly a dominant color.
- An interval [4, 4] contains one ball, with color 2 again.
- An interval [2, 4] contains two balls of color 2 and one ball of color 1.
There are 7 more intervals and color 1 is dominant in all of them.
Solution:
#include <bits/stdc++.h> using namespace std; const int N = 77777; int cnt[N], a[N], ans[N]; int main() { int n; scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%d", a + i); a[i]--; } for (int i = 0; i < n; i++) { ans[i] = 0; } for (int start = 0; start < n; start++) { for (int i = 0; i < n; i++) { cnt[i] = 0; } int best = 0; for (int i = start; i < n; i++) { cnt[a[i]]++; if (cnt[a[i]] > cnt[best] || (cnt[a[i]] == cnt[best] && a[i] < best)) { best = a[i]; } ans[best]++; } } for (int i = 0; i < n; i++) { if (i > 0) { putchar(' '); } printf("%d", ans[i]); } printf("\n"); return 0; }