#include <bits/stdc++.h>
using namespace std;
inline vector<vector<int> > merge(const vector<vector<int> >& a,
                                  const vector<vector<int> >& b) {
  if (a.empty()) return b;
  if (b.empty()) return a;
  vector<vector<int> > dp(5, vector<int>(5));
  for (int j = 0; j < 5; j++) {
    for (int i = 0; i <= j; i++) {
      unsigned int pb = UINT_MAX;
      for (int k = i; k <= j; k++) {
        if (~a[k][i] && ~b[j][k]) {
          pb = min(pb, (unsigned int)(a[k][i] + b[j][k]));
        }
      }
      dp[j][i] = pb;
    }
  }
  return dp;
}
inline void build(vector<vector<vector<int> > >& st, const int& n) {
  for (int i = n - 1; i > 0; i--) st[i] = merge(st[i << 1], st[(i << 1) + 1]);
}
inline int query(int& l, int& r, vector<vector<vector<int> > >& st,
                 const int& n) {
  vector<vector<int> > ansl, ansr;
  for (; l < r; l = l >> 1, r = r >> 1) {
    if (l & 1) {
      ansl = merge(ansl, st[l]);
      ++l;
    }
    if (r & 1) {
      --r;
      ansr = merge(st[r], ansr);
    }
  }
  return merge(ansl, ansr)[4][0];
}
int main() {
  int q, n;
  char s[200010];
  scanf("%d %d %s", &n, &q, s);
  vector<vector<vector<int> > > st(n << 1);
  for (int i = 0; i < n; i++) {
    switch (s[i]) {
      case '2':
        st[n + i] = {
            {1}, {0, 0}, {-1, -1, 0}, {-1, -1, -1, 0}, {-1, -1, -1, -1, 0}};
        break;
      case '0':
        st[n + i] = {
            {0}, {-1, 1}, {-1, 0, 0}, {-1, -1, -1, 0}, {-1, -1, -1, -1, 0}};
        break;
      case '1':
        st[n + i] = {
            {0}, {-1, 0}, {-1, -1, 1}, {-1, -1, 0, 0}, {-1, -1, -1, -1, 0}};
        break;
      case '7':
        st[n + i] = {
            {0}, {-1, 0}, {-1, -1, 0}, {-1, -1, -1, 1}, {-1, -1, -1, 0, 0}};
        break;
      case '6':
        st[n + i] = {
            {0}, {-1, 0}, {-1, -1, 0}, {-1, -1, -1, 1}, {-1, -1, -1, -1, 1}};
        break;
      default:
        st[n + i] = {
            {0}, {-1, 0}, {-1, -1, 0}, {-1, -1, -1, 0}, {-1, -1, -1, -1, 0}};
    }
  }
  build(st, n);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l += n - 1;
    r += n;
    printf("%d\n", query(l, r, st, n));
  }
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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;
inline vector<vector<int> > merge(const vector<vector<int> >& a,
                                  const vector<vector<int> >& b) {
  if (a.empty()) return b;
  if (b.empty()) return a;
  vector<vector<int> > dp(5, vector<int>(5));
  for (int j = 0; j < 5; j++) {
    for (int i = 0; i <= j; i++) {
      unsigned int pb = UINT_MAX;
      for (int k = i; k <= j; k++) {
        if (~a[k][i] && ~b[j][k]) {
          pb = min(pb, (unsigned int)(a[k][i] + b[j][k]));
        }
      }
      dp[j][i] = pb;
    }
  }
  return dp;
}
inline void build(vector<vector<vector<int> > >& st, const int& n) {
  for (int i = n - 1; i > 0; i--) st[i] = merge(st[i << 1], st[(i << 1) + 1]);
}
inline int query(int& l, int& r, vector<vector<vector<int> > >& st,
                 const int& n) {
  vector<vector<int> > ansl, ansr;
  for (; l < r; l = l >> 1, r = r >> 1) {
    if (l & 1) {
      ansl = merge(ansl, st[l]);
      ++l;
    }
    if (r & 1) {
      --r;
      ansr = merge(st[r], ansr);
    }
  }
  return merge(ansl, ansr)[4][0];
}
int main() {
  int q, n;
  char s[200010];
  scanf("%d %d %s", &n, &q, s);
  vector<vector<vector<int> > > st(n << 1);
  for (int i = 0; i < n; i++) {
    switch (s[i]) {
      case '2':
        st[n + i] = {
            {1}, {0, 0}, {-1, -1, 0}, {-1, -1, -1, 0}, {-1, -1, -1, -1, 0}};
        break;
      case '0':
        st[n + i] = {
            {0}, {-1, 1}, {-1, 0, 0}, {-1, -1, -1, 0}, {-1, -1, -1, -1, 0}};
        break;
      case '1':
        st[n + i] = {
            {0}, {-1, 0}, {-1, -1, 1}, {-1, -1, 0, 0}, {-1, -1, -1, -1, 0}};
        break;
      case '7':
        st[n + i] = {
            {0}, {-1, 0}, {-1, -1, 0}, {-1, -1, -1, 1}, {-1, -1, -1, 0, 0}};
        break;
      case '6':
        st[n + i] = {
            {0}, {-1, 0}, {-1, -1, 0}, {-1, -1, -1, 1}, {-1, -1, -1, -1, 1}};
        break;
      default:
        st[n + i] = {
            {0}, {-1, 0}, {-1, -1, 0}, {-1, -1, -1, 0}, {-1, -1, -1, -1, 0}};
    }
  }
  build(st, n);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l += n - 1;
    r += n;
    printf("%d\n", query(l, r, st, n));
  }
  return 0;
}

```