#include <bits/stdc++.h>
using namespace std;
string need = "2017#";
string big;
typedef int table[5][5];
table DP[2000000];
void combine(int a, int b, int to) {
  static table A, B;
  memcpy(A, DP[a], sizeof(A));
  memcpy(B, DP[b], sizeof(B));
  memset(DP[to], -1, sizeof(DP[to]));
  for (int i = 0; i < 5; ++i)
    for (int j = i; j < 5; ++j)
      for (int k = j; k < 5; ++k) {
        if (A[i][j] != -1 && B[j][k] != -1) {
          int now = A[i][j] + B[j][k];
          if (DP[to][i][k] == -1 || DP[to][i][k] > now) DP[to][i][k] = now;
        }
      }
}
void build_tree(int node, int b, int e) {
  if (b == e) {
    memset(DP[node], -1, sizeof(DP[node]));
    for (int i = 0; i < 5; ++i) {
      if (big[b] == need[i]) {
        DP[node][i][i] = 1;
        DP[node][i][i + 1] = 0;
      } else {
        DP[node][i][i] = (i >= 3 && big[b] == '6');
      }
    }
  } else {
    int m = (b + e) / 2;
    build_tree(node * 2, b, m);
    build_tree(node * 2 + 1, m + 1, e);
    combine(node * 2, node * 2 + 1, node);
  }
}
void query_tree(int node, int b, int e, int l, int r) {
  if (b > r || e < l) return;
  if (b >= l && e <= r) return combine(0, node, 0);
  int m = (b + e) / 2;
  query_tree(node * 2, b, m, l, r);
  query_tree(node * 2 + 1, m + 1, e, l, r);
}
void dump(table t) {
  for (int i = 0; i < 4; ++i)
    for (int j = i; j < 5; ++j) {
      cerr << i << " " << j << ": " << t[i][j] << endl;
    }
  cerr << endl << endl;
}
void dump_tree(int node, int b, int e) {
  cerr << "Dumping: " << node << "[" << b << " " << e << "]...\n";
  dump(DP[node]);
  if (b == e) return;
  int m = (b + e) / 2;
  dump_tree(node * 2, b, m);
  dump_tree(node * 2 + 1, m + 1, e);
}
int main() {
  int n, q;
  cin >> n >> q >> big;
  build_tree(1, 0, n - 1);
  while (q--) {
    int a, b;
    cin >> a >> b;
    memset(DP[0], -1, sizeof(DP[0]));
    for (int i = 0; i < 5; ++i) DP[0][i][i] = 0;
    query_tree(1, 0, n - 1, a - 1, b - 1);
    cout << DP[0][0][4] << '\n';
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

A string t is called nice if a string "2017" occurs in t as a subsequence but a string "2016" doesn't occur in t as a subsequence. For example, strings "203434107" and "9220617" are nice, while strings "20016", "1234" and "20167" aren't nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it's impossible to make a string nice by removing characters, its ugliness is  - 1.

Limak has a string s of length n, with characters indexed 1 through n. He asks you q queries. In the i-th query you should compute and print the ugliness of a substring (continuous subsequence) of s starting at the index ai and ending at the index bi (inclusive).

Input

The first line of the input contains two integers n and q (4 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the length of the string s and the number of queries respectively.

The second line contains a string s of length n. Every character is one of digits '0'–'9'.

The i-th of next q lines contains two integers ai and bi (1 ≤ ai ≤ bi ≤ n), describing a substring in the i-th query.

Output

For each query print the ugliness of the given substring.

Examples

Input

8 3
20166766
1 8
1 7
2 8


Output

4
3
-1


Input

15 5
012016662091670
3 4
1 14
4 15
1 13
10 15


Output

-1
2
1
-1
-1


Input

4 2
1234
2 4
1 2


Output

-1
-1

Note

In the first sample:

  * In the first query, ugliness("20166766") = 4 because all four sixes must be removed. 
  * In the second query, ugliness("2016676") = 3 because all three sixes must be removed. 
  * In the third query, ugliness("0166766") = - 1 because it's impossible to remove some digits to get a nice string. 



In the second sample:

  * In the second query, ugliness("01201666209167") = 2. It's optimal to remove the first digit '2' and the last digit '6', what gives a string "010166620917", which is nice. 
  * In the third query, ugliness("016662091670") = 1. It's optimal to remove the last digit '6', what gives a nice string "01666209170". 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
string need = "2017#";
string big;
typedef int table[5][5];
table DP[2000000];
void combine(int a, int b, int to) {
  static table A, B;
  memcpy(A, DP[a], sizeof(A));
  memcpy(B, DP[b], sizeof(B));
  memset(DP[to], -1, sizeof(DP[to]));
  for (int i = 0; i < 5; ++i)
    for (int j = i; j < 5; ++j)
      for (int k = j; k < 5; ++k) {
        if (A[i][j] != -1 && B[j][k] != -1) {
          int now = A[i][j] + B[j][k];
          if (DP[to][i][k] == -1 || DP[to][i][k] > now) DP[to][i][k] = now;
        }
      }
}
void build_tree(int node, int b, int e) {
  if (b == e) {
    memset(DP[node], -1, sizeof(DP[node]));
    for (int i = 0; i < 5; ++i) {
      if (big[b] == need[i]) {
        DP[node][i][i] = 1;
        DP[node][i][i + 1] = 0;
      } else {
        DP[node][i][i] = (i >= 3 && big[b] == '6');
      }
    }
  } else {
    int m = (b + e) / 2;
    build_tree(node * 2, b, m);
    build_tree(node * 2 + 1, m + 1, e);
    combine(node * 2, node * 2 + 1, node);
  }
}
void query_tree(int node, int b, int e, int l, int r) {
  if (b > r || e < l) return;
  if (b >= l && e <= r) return combine(0, node, 0);
  int m = (b + e) / 2;
  query_tree(node * 2, b, m, l, r);
  query_tree(node * 2 + 1, m + 1, e, l, r);
}
void dump(table t) {
  for (int i = 0; i < 4; ++i)
    for (int j = i; j < 5; ++j) {
      cerr << i << " " << j << ": " << t[i][j] << endl;
    }
  cerr << endl << endl;
}
void dump_tree(int node, int b, int e) {
  cerr << "Dumping: " << node << "[" << b << " " << e << "]...\n";
  dump(DP[node]);
  if (b == e) return;
  int m = (b + e) / 2;
  dump_tree(node * 2, b, m);
  dump_tree(node * 2 + 1, m + 1, e);
}
int main() {
  int n, q;
  cin >> n >> q >> big;
  build_tree(1, 0, n - 1);
  while (q--) {
    int a, b;
    cin >> a >> b;
    memset(DP[0], -1, sizeof(DP[0]));
    for (int i = 0; i < 5; ++i) DP[0][i][i] = 0;
    query_tree(1, 0, n - 1, a - 1, b - 1);
    cout << DP[0][0][4] << '\n';
  }
  return 0;
}

```