Given an array $a$ of $n$ integers and an integer $k$ ($2 \le k \le n$), where each element of the array is denoted by $a_i$ ($0 \le i < n$). Perform the operation $z$ given below on $a$ and print the value of $z(a,k)$ modulo $10^{9}+7$.
function z(array a, integer k):
if length(a) < k:
return 0
else:
b = empty array
ans = 0
for i = 0 .. (length(a) - k):
temp = a[i]
for j = i .. (i + k - 1):
temp = max(temp, a[j])
append temp to the end of b
ans = ans + temp
return ans + z(b, k)
Input
The first line of input contains two integers $n$ and $k$ ($2 \le k \le n \le 10^6$) — the length of the initial array $a$ and the parameter $k$.
The second line of input contains $n$ integers $a_0, a_1, \ldots, a_{n – 1}$ ($1 \le a_{i} \le 10^9$) — the elements of the array $a$.
Output
Output the only integer, the value of $z(a,k)$ modulo $10^9+7$.
Examples
input
3 2
9 1 10
output
29
input
5 3
5 8 7 1 9
output
34
Note
In the first example:
- for $a=(9,1,10)$, $ans=19$ and $b=(9,10)$,
- for $a=(9,10)$, $ans=10$ and $b=(10)$,
- for $a=(10)$, $ans=0$.
So the returned value is $19+10+0=29$.
In the second example:
- for $a=(5,8,7,1,9)$, $ans=25$ and $b=(8,8,9)$,
- for $a=(8,8,9)$, $ans=9$ and $b=(9)$,
- for $a=(9)$, $ans=0$.
So the returned value is $25+9+0=34$.
Solution:
#include <bits/stdc++.h> using namespace std; const int md = (int) 1e9 + 7; inline void add(int &a, int b) { a += b; if (a >= md) a -= md; } inline void sub(int &a, int b) { a -= b; if (a < 0) a += md; } inline int mul(int a, int b) { return (int) ((long long) a * b % md); } inline int power(int a, long long b) { int res = 1; while (b > 0) { if (b & 1) { res = mul(res, a); } a = mul(a, a); b >>= 1; } return res; } inline int inv(int a) { a %= md; if (a < 0) a += md; int b = md, u = 0, v = 1; while (a) { int t = b / a; b -= t * a; swap(a, b); u -= t * v; swap(u, v); } assert(b == 1); if (u < 0) u += md; return u; } const int INV2 = inv(2); inline int sum_positive(int start, int finish, int step) { if (start <= 0) { return 0; } if (finish < 0) { finish += abs(finish) / step * step; } while (finish < 0) { finish += step; } int cnt = (start - finish) / step + 1; int value = mul(mul(start + finish, cnt), INV2); return value; } int main() { ios::sync_with_stdio(false); cin.tie(0); int n, k; cin >> n >> k; vector<pair<int,int>> e(n); for (int i = 0; i < n; i++) { cin >> e[i].first; e[i].second = i; } sort(e.begin(), e.end()); vector<int> pr(n), ne(n); for (int i = 0; i < n; i++) { pr[i] = i - 1; ne[i] = i + 1; } const int step = k - 1; int ans = 0; for (auto &pp : e) { int i = pp.second; int L = pr[i] + 1; int R = ne[i] - 1; int len = (R - L + 1); if (len >= k) { int pos = i - L; int xstart = k; int xfinish = k + (len - k) / step * step; int xcnt = (xfinish - xstart) / step + 1; { int value = sum_positive(pos + 1 - xstart, pos + 1 - xfinish, step); sub(ans, mul(value, pp.first)); } { add(ans, mul(mul(pos, xcnt), pp.first)); int value = sum_positive(xfinish + pos - len, xstart + pos - len, step); sub(ans, mul(value, pp.first)); } add(ans, mul(xcnt, pp.first)); } if (pr[i] != -1) { ne[pr[i]] = ne[i]; } if (ne[i] != n) { pr[ne[i]] = pr[i]; } } cout << ans << '\n'; return 0; }