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>
const long long md = 1e9 + 7;
const int Inf = 1e9;
const long long Inf64 = 1e18;
const long long MaxN = 2e5 + 10;
const long long MaxM = 11;
const long double eps = 1e-15;
const long long dx[4] = {0, 1, 0, -1};
const long long dy[4] = {1, 0, -1, 0};
const long long ddx[4] = {1, 1, -1, -1};
const long long ddy[4] = {1, -1, 1, -1};
const long double Pi = acos(-1);
using namespace std;
long long gcd(long long a, long long b) {
  while (a) {
    b %= a;
    swap(a, b);
  }
  return b;
}
long long gcdex(long long a, long long mod = md) {
  long long g = mod, r = a, x = 0, y = 1;
  while (r != 0) {
    long long q = g / r;
    g %= r;
    swap(g, r);
    x -= q * y;
    swap(x, y);
  }
  return x < 0 ? x + mod : x;
}
long long binpow(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) {
      res *= a;
      res %= md;
    }
    a *= a;
    a %= md;
    n >>= 1;
  }
  return res % md;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  cout << fixed, cout.precision(10);
  vector<long long> St(1e5 + 1);
  St[0] = 1;
  for (int i = 1; i < 1e5 + 1; i++) {
    St[i] = St[i - 1] * 2;
    St[i] %= md;
  }
  long long N, Q;
  cin >> N >> Q;
  vector<pair<long long, long long> > Mt(N + 1);
  for (int i = 1; i <= N; i++) {
    char a;
    cin >> a;
    a -= '0';
    Mt[i].first = Mt[i - 1].first;
    Mt[i].second = Mt[i - 1].second;
    if (a) {
      Mt[i].second++;
    } else
      Mt[i].first++;
  }
  for (int i = 0; i < Q; i++) {
    long long a, b;
    cin >> a >> b;
    long long K = St[Mt[b].second - Mt[a - 1].second];
    K--;
    if (K < 0) K += md;
    K *= St[Mt[b].first - Mt[a - 1].first];
    K %= md;
    cout << K << '\n';
  }
  cerr << '\n'
       << "Time execute: " << clock() / (double)CLOCKS_PER_SEC << " sec"
       << '\n';
  return 0;
}