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;
inline long long int __gcd(long long int a, long long int b);
void pdash(int n = 1);
int bitcount(long long int u);
long long int power(long long int x, long long int y);
long long int power(long long int x, long long int y, long long int z);
long long int modInverse(long long int n, long long int p);
long long int nCrF(long long int n, long long int r, long long int p);
void cordinate_compression(vector<int>& v);
void make_unique(vector<int>& vec);
const long long int mod = 1e9 + 7;
void solve() {
  int n, q;
  cin >> n >> q;
  string str;
  cin >> str;
  vector<int> vec(n);
  for (int i = 0; i < n; i++) {
    vec[i] = (str[i] - '0');
  }
  for (int i = 1; i < n; i++) {
    vec[i] += vec[i - 1];
  }
  while (q--) {
    int zero, one;
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    one = vec[b] - (((a - 1) >= 0) ? vec[a - 1] : 0);
    zero = (b - a) + 1 - one;
    cout << ((power(2, zero, mod) * ((power(2, one, mod) - 1 + mod) % mod)) %
             mod)
         << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}
inline long long int __gcd(long long int a, long long int b) {
  if (a == 0 || b == 0) {
    return max(a, b);
  }
  long long int tempa, tempb;
  while (1) {
    if (a % b == 0)
      return b;
    else {
      tempa = a;
      tempb = b;
      a = tempb;
      b = tempa % tempb;
    }
  }
}
void pdash(int n) {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 30; j++) {
      cout << "-";
    }
    cout << "\n";
  }
}
long long int power(long long int x, long long int y) {
  long long int result = 1;
  while (y > 0) {
    if (y & 1) {
      result = (result * x);
    }
    y = y >> 1;
    x = (x * x);
  }
  return result;
}
long long int power(long long int x, long long int y, long long int z) {
  long long int result = 1;
  x = x % z;
  while (y > 0) {
    if (y & 1) {
      result = (result * x) % z;
    }
    y = y >> 1;
    x = (x * x) % z;
  }
  return result;
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
long long int nCrF(long long int n, long long int r, long long int p) {
  if (r == 0) return 1;
  long long int f[n + 1];
  f[0] = 1;
  for (long long int i = 1; i <= n; i++) f[i] = f[i - 1] * i % p;
  return (f[n] * modInverse(f[r], p) % p * modInverse(f[n - r], p) % p) % p;
}
void cordinate_compression(vector<int>& v) {
  vector<int> p = v;
  make_unique(p);
  for (int i = 0; i < (int)((v).size()); i++)
    v[i] = (int)(lower_bound(p.begin(), p.end(), v[i]) - p.begin());
}
void make_unique(vector<int>& vec) {
  sort(vec.begin(), vec.end());
  vec.erase(unique(vec.begin(), vec.end()), vec.end());
}
int bitcount(long long int u) {
  int cnt = 0;
  while (u) {
    u = u & (u - 1);
    cnt++;
  }
  return cnt;
}