Problem: JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.

First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.

For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.

However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.

All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).

The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.

Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.

Output

Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.

Examples

Input

4 2
1011
1 4
3 4


Output

14
3


Input

3 2
111
1 2
3 3


Output

3
1

Note

In the first example: 

  * For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2. 
  * For query 2: Both 3, 4 and 4, 3 ordering give the same answer. 



In the second example, any order of eating parts leads to the same answer.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void codeforces_520_A() {
  int n;
  scanf("%d", &n);
  int a[n];
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  int ans = 0;
  if (n == 1) {
    printf("0");
    return;
  } else if (n == 2) {
    if (a[0] == 1 && a[1] == 2) {
      printf("1");
    } else if (a[0] == 999 && a[1] == 1000) {
      printf("1");
    } else {
      printf("0");
    }
    return;
  }
  int ansans = 0;
  if (a[0] == 1 && a[1] == 2) {
    ansans++;
  }
  for (int i = 1; i < n - 1; i++) {
    if (a[i] == a[i - 1] + 1 && a[i] == a[i + 1] - 1) {
      ansans++;
    } else {
      ans = max(ans, ansans);
      ansans = 0;
    }
  }
  if (a[n - 2] == 999 && a[n - 1] == 1000) {
    ansans++;
  }
  ans = max(ans, ansans);
  printf("%d", ans);
}
void codeforces_520_B() {
  int prime[1000001] = {0};
  prime[0] = 1;
  prime[1] = 1;
  for (int i = 2; i < 1000001; i++) {
    if (prime[i] == 0) {
      for (int j = i; j < 1000001; j += i) {
        prime[j] = 1;
      }
      prime[i] = 0;
    }
  }
  vector<int> primes;
  for (int i = 0; i < 10001; i++) {
    if (prime[i] == 0) {
      primes.push_back(i);
    }
  }
  int n;
  scanf("%d", &n);
  if (n == 1) {
    printf("1 0");
    return;
  }
  vector<pair<int, int> > howmanyprime;
  int cur = 0;
  int num = 0;
  while (n != 1) {
    if (n % primes[cur] == 0) {
      n /= primes[cur];
      num++;
    } else if (n % primes[cur] != 0) {
      if (num != 0) {
        howmanyprime.push_back({primes[cur], num});
        num = 0;
      }
      cur++;
      if (primes[cur] * primes[cur] > n) {
        cur = 10000000;
        break;
      }
    }
  }
  if (cur == 10000000) {
    howmanyprime.push_back({n, 1});
  }
  if (num != 0) {
    howmanyprime.push_back({primes[cur], num});
  }
  int realans = 1;
  for (int i = 0; i < howmanyprime.size(); i++) {
    realans *= howmanyprime[i].first;
  }
  int realtime = 0;
  int mmax = 0;
  int isthereodd = 0;
  for (int i = 0; i < howmanyprime.size(); i++) {
    if (mmax < howmanyprime[i].second) {
      mmax = howmanyprime[i].second;
    }
  }
  for (int i = 0; i < howmanyprime.size(); i++) {
    if (howmanyprime[i].second < mmax) {
      isthereodd = 1;
      break;
    }
  }
  if ((mmax != 1) && (mmax != 2) && (mmax != 4) && (mmax != 8) &&
      (mmax != 16) && (mmax != 32)) {
    isthereodd = 1;
  }
  realtime += isthereodd;
  while (mmax != 1) {
    mmax = (mmax + 1) / 2;
    realtime++;
  }
  printf("%d %d", realans, realtime);
}
void codeforces_520_C() {
  int pow2[1000005];
  pow2[0] = 1;
  pow2[1] = 2;
  for (int i = 2; i < 1000005; i++) {
    pow2[i] = (pow2[i - 1] * 2) % 1000000007;
  }
  int n, q;
  scanf("%d %d", &n, &q);
  char s[n + 5];
  scanf("%s", s);
  int good[n + 3];
  for (int i = 1; i <= n; i++) {
    good[i] = s[i - 1] - '0';
  }
  int howmanyzero[n + 3];
  int howmanyone[n + 3];
  howmanyzero[0] = 0;
  howmanyone[0] = 0;
  for (int i = 1; i <= n; i++) {
    if (good[i] == 1) {
      howmanyone[i] = howmanyone[i - 1] + 1;
      howmanyzero[i] = howmanyzero[i - 1];
    } else {
      howmanyone[i] = howmanyone[i - 1];
      howmanyzero[i] = howmanyzero[i - 1] + 1;
    }
  }
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    int numone = howmanyone[r] - howmanyone[l - 1];
    int numzero = howmanyzero[r] - howmanyzero[l - 1];
    printf("%d\n",
           (pow2[numone + numzero] - pow2[numzero] + 1000000007) % 1000000007);
  }
}
void codeforces_520_D() {}
void codeforces_520_E() {}
void codeforces_520_F() {}
int main(void) {
  codeforces_520_C();
  codeforces_520_D();
  codeforces_520_E();
  codeforces_520_F();
  return 0;
}