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.
/*
  *  sort(array_name,array_name + array_size);                  // to sort the array in ascending order
  *  sort(array_name,array_name + array_size,greater<int>());  // to sort the array in descending order
  *  sort(v.begin(),v.end());                      //  to sort the vector in ascending order
  *  sort(array_name + starting index,array_name + desired index+1)  //specific sorting (similary for vector)
  *  __gcd();                                     // to find gcd of two numbers
  *  memset(array_name,num,sizeof(array_name));  // num= -1 OR 0 to set the entire array equal num
  *  __builtin_popcount(num)                    //  to count no. of 1's in num
  *  string s; cin>>s; stops reading after a space so use getline(cin,s);
     getline(cin,s,'x'); will stop after reading 'x'(also wont read x) or any char value given
  *  int m=*max_element(a,a+n);                  // gives max element in the array
  *  int m=*max_element(v.begin(),v.end());      // gives max element in the vector
  *  s.find(str); gives the index of the first occurence of the given string str
  *  s=to_string(x);            //int x; string s;
  *  x=stoi(s);                 //int x; string s;
  ASCII Value of A=65   Z=90    a=97    z=122   0=48    9=57 
*/
 
#include<bits/stdc++.h>
using namespace std;
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define ll long long
#define fr(i, n)     for(ll i=0;i<n;i++)
#define ffr(i,a,b)   for(ll i=a;i<=b;i++)
#define bfr(i,a,b)   for(ll i=a;i>=b;i--)
#define fio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define tst() ll t; cin>>t; while(t--)
ll gcd(ll x,ll y){ if(x==0)return y; return(gcd(y%x,x)); } //__gcd(); #include<algorithm>
string Tolower(string s) { transform(s.begin(), s.end(), s.begin(), ::tolower); return s; }
string Toupper(string s) { transform(s.begin(), s.end(), s.begin(), ::toupper); return s; }
bool prime[100001];
void sieve()
{
    memset(prime,true,sizeof(prime));
    prime[0]=prime[1]=false;
    for(ll i=2;i*i<100001;i++)
    if(prime[i]==true)
    for(ll j=i*i;j<100001;j+=i)
    prime[j]=false;
    return;
}

vector<ll> primef(5000001,-1);
void primefact()
{
    for(ll i=2;i<5000001;i++)
    {
        if(primef[i]==-1)
        {
            primef[i]=i;
            for(ll j=i*i;j<5000001;j+=i)
            if(primef[j]==-1)
            primef[j]=i;
        }
    }
    return ;
}

bool lenaplcheck(pair<string,int> &a,pair<string,int> &b)
{
    return (a.first).size() < (b.first).size();
}   // to sort pair<string,int> acsending order of length first then alpha

bool mycomp(pair<ll,ll> &p,pair<ll,ll> &p1)
{
    double a,b;
    a=(double)p.second/(double)p.first;
    b=(double)p1.second/(double)p1.first;
    return (a<b);
}


void solve()
{
    ll n,q;
    cin>>n>>q;
    ll a[n+1];
    string s;
    cin>>s;
    ffr(i,0,n-1)
    a[i+1]=s[i]-'a'+1;
    ffr(i,1,n)
    a[i]+=a[i-1];
    while(q--)
    {
        ll l,r;
        cin>>l>>r;
        cout<<a[r]-a[l-1]<<"\n";
    }
    return;
}
int main()
{
	fio;
    solve();
    return 0;
}