#include <bits/stdc++.h>
using namespace std;
class Solver {
 public:
  static const int maxn = 220000;
  char str[maxn];
  class Seg {
   public:
    int v[6][6];
    Seg operator+(const Seg& b) const {
      Seg tmp;
      for (int i = 0; i < 5; i++)
        for (int j = 0; j < 5; j++) tmp.v[i][j] = maxn;
      for (int i = 0; i < 5; i++)
        for (int j = 0; j < 5; j++)
          for (int k = 0; k < 5; k++)
            tmp.v[i][j] = min(tmp.v[i][j], v[i][k] + b.v[k][j]);
      return tmp;
    }
  } tr[maxn << 2];
  void pushup(int rt) { tr[rt] = tr[rt << 1] + tr[rt << 1 | 1]; }
  void init(Seg& tt) {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) tt.v[i][j] = (i == j ? 0 : maxn);
  }
  void show(Seg x) {
    for (int i = 0; i < 5; i++, puts(""))
      for (int j = 0; j < 5; j++) printf("%d ", x.v[i][j]);
  }
  void build(int l, int r, int rt) {
    init(tr[rt]);
    if (l == r) {
      if (str[l] == '2') tr[rt].v[0][1] = 0, tr[rt].v[0][0] = 1;
      if (str[l] == '0') tr[rt].v[1][2] = 0, tr[rt].v[1][1] = 1;
      if (str[l] == '1') tr[rt].v[2][3] = 0, tr[rt].v[2][2] = 1;
      if (str[l] == '7') tr[rt].v[3][4] = 0, tr[rt].v[3][3] = 1;
      if (str[l] == '6') tr[rt].v[3][3] = 1, tr[rt].v[4][4] = 1;
      return;
    }
    int m = (l + r) >> 1;
    if (l <= m) build(l, m, rt << 1);
    if (r > m) build(m + 1, r, rt << 1 | 1);
    pushup(rt);
  }
  int n, q;
  Seg ask(int L, int R, int l, int r, int rt) {
    if (L <= l && R >= r) {
      return tr[rt];
    }
    int m = (l + r) >> 1;
    Seg tmp;
    init(tmp);
    if (L <= m) tmp = tmp + ask(L, R, l, m, rt << 1);
    if (R > m) tmp = tmp + ask(L, R, m + 1, r, rt << 1 | 1);
    return tmp;
  }
  void solve(int _n, int _q) {
    n = _n;
    q = _q;
    str[0] = 'o';
    scanf("%s", str + 1);
    build(1, n, 1);
    for (int i = 0; i < q; i++) {
      int l, r;
      scanf("%d%d", &l, &r);
      int ans = ask(l, r, 1, n, 1).v[0][4];
      if (ans >= maxn)
        puts("-1");
      else
        printf("%d\n", ans);
    }
  }
} solver;
int main() {
  int n, q;
  while (~scanf("%d%d", &n, &q)) solver.solve(n, q);
  return 0;
}

### Prompt

Your task is to create 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;
class Solver {
 public:
  static const int maxn = 220000;
  char str[maxn];
  class Seg {
   public:
    int v[6][6];
    Seg operator+(const Seg& b) const {
      Seg tmp;
      for (int i = 0; i < 5; i++)
        for (int j = 0; j < 5; j++) tmp.v[i][j] = maxn;
      for (int i = 0; i < 5; i++)
        for (int j = 0; j < 5; j++)
          for (int k = 0; k < 5; k++)
            tmp.v[i][j] = min(tmp.v[i][j], v[i][k] + b.v[k][j]);
      return tmp;
    }
  } tr[maxn << 2];
  void pushup(int rt) { tr[rt] = tr[rt << 1] + tr[rt << 1 | 1]; }
  void init(Seg& tt) {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) tt.v[i][j] = (i == j ? 0 : maxn);
  }
  void show(Seg x) {
    for (int i = 0; i < 5; i++, puts(""))
      for (int j = 0; j < 5; j++) printf("%d ", x.v[i][j]);
  }
  void build(int l, int r, int rt) {
    init(tr[rt]);
    if (l == r) {
      if (str[l] == '2') tr[rt].v[0][1] = 0, tr[rt].v[0][0] = 1;
      if (str[l] == '0') tr[rt].v[1][2] = 0, tr[rt].v[1][1] = 1;
      if (str[l] == '1') tr[rt].v[2][3] = 0, tr[rt].v[2][2] = 1;
      if (str[l] == '7') tr[rt].v[3][4] = 0, tr[rt].v[3][3] = 1;
      if (str[l] == '6') tr[rt].v[3][3] = 1, tr[rt].v[4][4] = 1;
      return;
    }
    int m = (l + r) >> 1;
    if (l <= m) build(l, m, rt << 1);
    if (r > m) build(m + 1, r, rt << 1 | 1);
    pushup(rt);
  }
  int n, q;
  Seg ask(int L, int R, int l, int r, int rt) {
    if (L <= l && R >= r) {
      return tr[rt];
    }
    int m = (l + r) >> 1;
    Seg tmp;
    init(tmp);
    if (L <= m) tmp = tmp + ask(L, R, l, m, rt << 1);
    if (R > m) tmp = tmp + ask(L, R, m + 1, r, rt << 1 | 1);
    return tmp;
  }
  void solve(int _n, int _q) {
    n = _n;
    q = _q;
    str[0] = 'o';
    scanf("%s", str + 1);
    build(1, n, 1);
    for (int i = 0; i < q; i++) {
      int l, r;
      scanf("%d%d", &l, &r);
      int ans = ask(l, r, 1, n, 1).v[0][4];
      if (ans >= maxn)
        puts("-1");
      else
        printf("%d\n", ans);
    }
  }
} solver;
int main() {
  int n, q;
  while (~scanf("%d%d", &n, &q)) solver.solve(n, q);
  return 0;
}

```