Problem: Petya once wrote a sad love song and shared it to Vasya. The song is a string consisting of lowercase English letters. Vasya made up q questions about this song. Each question is about a subsegment of the song starting from the l-th letter to the r-th letter. Vasya considers a substring made up from characters on this segment and repeats each letter in the subsegment k times, where k is the index of the corresponding letter in the alphabet. For example, if the question is about the substring "abbcb", then Vasya repeats letter 'a' once, each of the letters 'b' twice, letter 'c" three times, so that the resulting string is "abbbbcccbb", its length is 10. Vasya is interested about the length of the resulting string.

Help Petya find the length of each string obtained by Vasya.

Input

The first line contains two integers n and q (1≤ n≤ 100 000, 1≤ q ≤ 100 000) — the length of the song and the number of questions. 

The second line contains one string s — the song, consisting of n lowercase letters of English letters.

Vasya's questions are contained in the next q lines. Each line contains two integers l and r (1 ≤ l ≤ r ≤ n) — the bounds of the question.

Output

Print q lines: for each question print the length of the string obtained by Vasya.

Examples

Input


7 3
abacaba
1 3
2 5
1 7


Output


4
7
11


Input


7 4
abbabaa
1 3
5 7
6 6
2 4


Output


5
4
1
5


Input


13 7
sonoshikumiwo
1 5
2 10
7 7
1 13
4 8
2 5
3 9


Output


82
125
9
191
62
63
97

Note

In the first example Vasya is interested in three questions. In the first question Vasya considers the substring "aba", that transforms to "abba", so the answer is equal to 4. In the second question Vasya considers "baca", that transforms to "bbaccca", so the answer is 7. In the third question Vasya considers the string "abacaba",that transforms to "abbacccabba" of length 11.

Solution (in CPP):

Solution:
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <vector>
#include <bitset>
#include <string>
#include <cstring>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <utility>
#include <algorithm>
#include <random>
#include <cmath>
#include <cassert>
#include <climits>
#include <ctime>
#include <chrono>


/*
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,popcnt,abm,mmx,avx,avx2,tune=native")
*/


#ifdef LOCAL
    #define dbg(x) cout << #x << " : " << x << endl;
#else
    #define dbg(x)
#endif

#define int long long
#define pb push_back
#define ppb pop_back()
#define mp make_pair
#define fi(a, b) for (int i = a; i < b; i++)
#define fj(a, b) for (int j = a; j < b; j++)
#define fk(a, b) for (int k = a; k < b; k++)
#define fi1(a, b) for (int i = a - 1; i >= b; i--)
#define fj1(a, b) for (int j = a - 1; j >= b; j--)
#define fk1(a, b) for (int k = a - 1; k >= b; k--)
#define fx(x, a) for (auto& x : a)
#define rep(i, a, b) for (int i = a; i < b; ++i)
#define rep1(i, a, b) for (int i = a - 1; i >= b; --i)
#define siz(x) (int)x.size()
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define left leftkekmda
#define right rightkekmda

using namespace std;

template<typename T1, typename T2>inline void mine(T1 &x, const T2 &y) { if (y < x) x = y; }
template<typename T1, typename T2>inline void maxe(T1 &x, const T2 &y) { if (x < y) x = y; }

template<typename T> ostream& operator << (ostream &os, const vector<T> &b) {
    for (auto &k : b) os << k << ' ';
    return os;
}
template<typename T> istream& operator >> (istream &is, vector<T> &b) {
    for (auto &k : b) is >> k;
    return is;
}


typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef char ch;
typedef string str;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpii;
typedef vector<vpii> vvpii;
typedef vector<ch> vch;
typedef vector<vch> vvch;
typedef vector<str> vs;


mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());


const int N = 1e5 + 10;
int ps[N][26];

int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int n, q;
    cin >> n >> q;
    string s;
    cin >> s;
    for (int i = 0; i < n; ++i) {
        s[i] -= 'a';
        for (int j = 0; j < 26; ++j) {
            ps[i + 1][j] = ps[i][j];
        }
        ps[i + 1][s[i]]++;
    }
    while (q--) {
        int l, r;
        cin >> l >> r;
        --l;
        int ans = 0;
        for (int j = 0; j < 26; ++j) {
            ans += (j + 1) * (ps[r][j] - ps[l][j]);
        }
        cout << ans << '\n';
    }
}