#include <bits/stdc++.h>
using namespace std;
inline char gc() {
  static char buf[1 << 16], *S, *T;
  if (S == T) {
    T = (S = buf) + fread(buf, 1, 1 << 16, stdin);
    if (T == S) return EOF;
  }
  return *S++;
}
inline int read() {
  int x = 0, f = 1;
  char ch = gc();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = gc();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = gc();
  return x * f;
}
char s[200010];
inline void get_S() {
  char ch = gc();
  int tot = 0;
  while (ch < '0' || ch > '9') ch = gc();
  while (ch >= '0' && ch <= '9') s[++tot] = ch, ch = gc();
}
int n, m;
struct node {
  int a[5][5];
  int* operator[](int x) { return a[x]; }
  inline void init() { memset(a, 0x3f3f3f3f, sizeof(a)); }
  friend node operator+(node a, node b) {
    node res;
    res.init();
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j)
        for (int k = 0; k < 5; ++k)
          res[i][j] = min(res[i][j], a[i][k] + b[k][j]);
    return res;
  }
} tr[200010 << 2];
inline void build(int p, int l, int r) {
  if (l == r) {
    tr[p].init();
    for (int i = 0; i < 5; ++i) tr[p][i][i] = 0;
    if (s[l] == '2') {
      tr[p][0][1] = 0;
      tr[p][0][0] = 1;
      return;
    }
    if (s[l] == '0') {
      tr[p][1][2] = 0;
      tr[p][1][1] = 1;
      return;
    }
    if (s[l] == '1') {
      tr[p][2][3] = 0;
      tr[p][2][2] = 1;
      return;
    }
    if (s[l] == '6') {
      tr[p][3][3] = 1;
      tr[p][4][4] = 1;
      return;
    }
    if (s[l] == '7') {
      tr[p][3][4] = 0;
      tr[p][3][3] = 0;
      return;
    }
    return;
  }
  int mid = l + r >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  tr[p] = tr[p << 1] + tr[p << 1 | 1];
}
inline node ask(int p, int l, int r, int x, int y) {
  if (x == l && r == y) return tr[p];
  int mid = l + r >> 1;
  if (y <= mid) return ask(p << 1, l, mid, x, y);
  if (x > mid) return ask(p << 1 | 1, mid + 1, r, x, y);
  return ask(p << 1, l, mid, x, mid) + ask(p << 1 | 1, mid + 1, r, mid + 1, y);
}
int main() {
  n = read();
  m = read();
  get_S();
  build(1, 1, n);
  while (m--) {
    int x = read(), y = read();
    int res = ask(1, 1, n, x, y).a[0][4];
    if (res == 0x3f3f3f3f)
      puts("-1");
    else
      printf("%d\n", res);
  }
  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 char gc() {
  static char buf[1 << 16], *S, *T;
  if (S == T) {
    T = (S = buf) + fread(buf, 1, 1 << 16, stdin);
    if (T == S) return EOF;
  }
  return *S++;
}
inline int read() {
  int x = 0, f = 1;
  char ch = gc();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = gc();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = gc();
  return x * f;
}
char s[200010];
inline void get_S() {
  char ch = gc();
  int tot = 0;
  while (ch < '0' || ch > '9') ch = gc();
  while (ch >= '0' && ch <= '9') s[++tot] = ch, ch = gc();
}
int n, m;
struct node {
  int a[5][5];
  int* operator[](int x) { return a[x]; }
  inline void init() { memset(a, 0x3f3f3f3f, sizeof(a)); }
  friend node operator+(node a, node b) {
    node res;
    res.init();
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j)
        for (int k = 0; k < 5; ++k)
          res[i][j] = min(res[i][j], a[i][k] + b[k][j]);
    return res;
  }
} tr[200010 << 2];
inline void build(int p, int l, int r) {
  if (l == r) {
    tr[p].init();
    for (int i = 0; i < 5; ++i) tr[p][i][i] = 0;
    if (s[l] == '2') {
      tr[p][0][1] = 0;
      tr[p][0][0] = 1;
      return;
    }
    if (s[l] == '0') {
      tr[p][1][2] = 0;
      tr[p][1][1] = 1;
      return;
    }
    if (s[l] == '1') {
      tr[p][2][3] = 0;
      tr[p][2][2] = 1;
      return;
    }
    if (s[l] == '6') {
      tr[p][3][3] = 1;
      tr[p][4][4] = 1;
      return;
    }
    if (s[l] == '7') {
      tr[p][3][4] = 0;
      tr[p][3][3] = 0;
      return;
    }
    return;
  }
  int mid = l + r >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  tr[p] = tr[p << 1] + tr[p << 1 | 1];
}
inline node ask(int p, int l, int r, int x, int y) {
  if (x == l && r == y) return tr[p];
  int mid = l + r >> 1;
  if (y <= mid) return ask(p << 1, l, mid, x, y);
  if (x > mid) return ask(p << 1 | 1, mid + 1, r, x, y);
  return ask(p << 1, l, mid, x, mid) + ask(p << 1 | 1, mid + 1, r, mid + 1, y);
}
int main() {
  n = read();
  m = read();
  get_S();
  build(1, 1, n);
  while (m--) {
    int x = read(), y = read();
    int res = ask(1, 1, n, x, y).a[0][4];
    if (res == 0x3f3f3f3f)
      puts("-1");
    else
      printf("%d\n", res);
  }
  return 0;
}

```