#include <bits/stdc++.h>
using namespace std;
string s;
struct node {
  int c[5][5];
} aint[804040];
int nums[] = {2, 0, 1, 6};
node comb(node a, node b) {
  node ret;
  for (int i = 0; i < (5); ++i)
    for (int j = 0; j < (5); ++j) ret.c[i][j] = -1;
  for (int i = 0; i < (5); ++i) {
    for (int j = i; j < (5); ++j) {
      for (int k = i; k < (j + 1); ++k) {
        if (a.c[i][k] != -1 && b.c[k][j] != -1) {
          if (ret.c[i][j] == -1) ret.c[i][j] = a.c[i][k] + b.c[k][j];
          ret.c[i][j] = min(a.c[i][k] + b.c[k][j], ret.c[i][j]);
        }
      }
    }
  }
  return ret;
}
void build(int nod, int st, int dr) {
  if (st == dr) {
    for (int i = 0; i < (5); ++i)
      for (int j = 0; j < (5); ++j) aint[nod].c[i][j] = -1;
    for (int i = 0; i < (4); ++i) {
      aint[nod].c[i][i] = 0;
      if (nums[i] == s[st - 1] - '0') {
        aint[nod].c[i][i] = 1;
        aint[nod].c[i][i + 1] = 0;
      }
    }
    return;
  }
  int mij = (st + dr) / 2;
  build(2 * nod, st, mij);
  build(2 * nod + 1, mij + 1, dr);
  aint[nod] = comb(aint[nod * 2], aint[2 * nod + 1]);
}
node quer(int nod, int st, int dr, int ist, int idr) {
  if (st >= ist && dr <= idr) {
    return aint[nod];
  } else {
    int mij = (st + dr) / 2;
    if (ist <= mij && idr > mij) {
      return comb(quer(2 * nod, st, mij, ist, idr),
                  quer(2 * nod + 1, mij + 1, dr, ist, idr));
    } else if (ist <= mij) {
      return quer(2 * nod, st, mij, ist, idr);
    } else {
      return quer(2 * nod + 1, mij + 1, dr, ist, idr);
    }
  }
}
int su[202020][10];
int bk[202020][10];
int N, M, L, R;
int sum(int st, int dr, int val) {
  --st;
  --dr;
  if (st == 0) return su[dr][val];
  return su[dr][val] - su[st - 1][val];
}
int main() {
  cin >> N >> M;
  cin >> s;
  build(1, 1, N);
  for (int i = 0; i < (10); ++i) {
    bk[0][i] = -1;
  }
  bk[0][s[0] - '0'] = 0;
  su[0][s[0] - '0']++;
  for (int i = 1; i < (N); ++i) {
    for (int j = 0; j < (10); ++j) {
      bk[i][j] = bk[i - 1][j];
      su[i][j] += su[i - 1][j];
    }
    su[i][s[i] - '0']++;
    bk[i][s[i] - '0'] = i;
  }
  for (int kx = 0; kx < (M); ++kx) {
    int x, y;
    cin >> x >> y;
    int oy = y;
    y = bk[y - 1][7] + 1;
    if (x > y) {
      cout << -1 << "\n";
      continue;
    }
    node c = quer(1, 1, N, x, y);
    if (c.c[0][3] == -1) {
      cout << -1 << "\n";
    } else {
      cout << c.c[0][3] + sum(y, oy, 6) << "\n";
    }
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 s;
struct node {
  int c[5][5];
} aint[804040];
int nums[] = {2, 0, 1, 6};
node comb(node a, node b) {
  node ret;
  for (int i = 0; i < (5); ++i)
    for (int j = 0; j < (5); ++j) ret.c[i][j] = -1;
  for (int i = 0; i < (5); ++i) {
    for (int j = i; j < (5); ++j) {
      for (int k = i; k < (j + 1); ++k) {
        if (a.c[i][k] != -1 && b.c[k][j] != -1) {
          if (ret.c[i][j] == -1) ret.c[i][j] = a.c[i][k] + b.c[k][j];
          ret.c[i][j] = min(a.c[i][k] + b.c[k][j], ret.c[i][j]);
        }
      }
    }
  }
  return ret;
}
void build(int nod, int st, int dr) {
  if (st == dr) {
    for (int i = 0; i < (5); ++i)
      for (int j = 0; j < (5); ++j) aint[nod].c[i][j] = -1;
    for (int i = 0; i < (4); ++i) {
      aint[nod].c[i][i] = 0;
      if (nums[i] == s[st - 1] - '0') {
        aint[nod].c[i][i] = 1;
        aint[nod].c[i][i + 1] = 0;
      }
    }
    return;
  }
  int mij = (st + dr) / 2;
  build(2 * nod, st, mij);
  build(2 * nod + 1, mij + 1, dr);
  aint[nod] = comb(aint[nod * 2], aint[2 * nod + 1]);
}
node quer(int nod, int st, int dr, int ist, int idr) {
  if (st >= ist && dr <= idr) {
    return aint[nod];
  } else {
    int mij = (st + dr) / 2;
    if (ist <= mij && idr > mij) {
      return comb(quer(2 * nod, st, mij, ist, idr),
                  quer(2 * nod + 1, mij + 1, dr, ist, idr));
    } else if (ist <= mij) {
      return quer(2 * nod, st, mij, ist, idr);
    } else {
      return quer(2 * nod + 1, mij + 1, dr, ist, idr);
    }
  }
}
int su[202020][10];
int bk[202020][10];
int N, M, L, R;
int sum(int st, int dr, int val) {
  --st;
  --dr;
  if (st == 0) return su[dr][val];
  return su[dr][val] - su[st - 1][val];
}
int main() {
  cin >> N >> M;
  cin >> s;
  build(1, 1, N);
  for (int i = 0; i < (10); ++i) {
    bk[0][i] = -1;
  }
  bk[0][s[0] - '0'] = 0;
  su[0][s[0] - '0']++;
  for (int i = 1; i < (N); ++i) {
    for (int j = 0; j < (10); ++j) {
      bk[i][j] = bk[i - 1][j];
      su[i][j] += su[i - 1][j];
    }
    su[i][s[i] - '0']++;
    bk[i][s[i] - '0'] = i;
  }
  for (int kx = 0; kx < (M); ++kx) {
    int x, y;
    cin >> x >> y;
    int oy = y;
    y = bk[y - 1][7] + 1;
    if (x > y) {
      cout << -1 << "\n";
      continue;
    }
    node c = quer(1, 1, N, x, y);
    if (c.c[0][3] == -1) {
      cout << -1 << "\n";
    } else {
      cout << c.c[0][3] + sum(y, oy, 6) << "\n";
    }
  }
}

```