#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
const int INF = 1 << 29;
const int MOD = 1e9 + 7;
const int N = 2e5 + 10;
const int K = 5;
char s[N];
struct Mat {
  int a[5][5];
};
Mat operator+(const Mat& a, const Mat& b) {
  Mat ret;
  for (int i = 0; i < K; ++i) {
    for (int j = 0; j < K; ++j) {
      ret.a[i][j] = INF;
      for (int k = 0; k < K; ++k) {
        ret.a[i][j] = min(ret.a[i][j], a.a[i][k] + b.a[k][j]);
      }
    }
  }
  return ret;
}
struct Node {
  int a, b;
  Node *left, *right;
  Mat sum;
};
Node pool[N << 1], *last;
Node* build(int a, int b) {
  Node* cur = last++;
  cur->a = a;
  cur->b = b;
  if (a + 1 == b) {
    for (int i = 0; i < K; ++i) {
      for (int j = 0; j < K; ++j) {
        cur->sum.a[i][j] = INF;
      }
      cur->sum.a[i][i] = 0;
    }
    if (s[a] == '2') {
      cur->sum.a[0][0] = 1;
      cur->sum.a[0][1] = 0;
    } else if (s[a] == '0') {
      cur->sum.a[1][1] = 1;
      cur->sum.a[1][2] = 0;
    } else if (s[a] == '1') {
      cur->sum.a[2][2] = 1;
      cur->sum.a[2][3] = 0;
    } else if (s[a] == '6') {
      cur->sum.a[3][3] = 1;
      cur->sum.a[4][4] = 1;
    } else if (s[a] == '7') {
      cur->sum.a[3][3] = 1;
      cur->sum.a[3][4] = 0;
    }
    return cur;
  }
  cur->left = build(a, (a + b) / 2);
  cur->right = build((a + b) / 2, b);
  cur->sum = cur->left->sum + cur->right->sum;
  return cur;
}
int A, B;
Mat query(Node* cur) {
  if (cur->a >= A && cur->b <= B) {
    return cur->sum;
  }
  Mat ret;
  for (int i = 0; i < K; ++i) {
    for (int j = 0; j < K; ++j) {
      ret.a[i][j] = INF;
    }
    ret.a[i][i] = 0;
  }
  if ((cur->a + cur->b) / 2 > A) ret = ret + query(cur->left);
  if ((cur->a + cur->b) / 2 < B) ret = ret + query(cur->right);
  return ret;
}
void debug(Node* cur, int dep) {
  if (cur == nullptr) return;
  for (int i = 0; i < dep; ++i) putchar(' ');
  printf("[%d,%d):\n", cur->a, cur->b);
  for (int i = 0; i < K; ++i) {
    for (int j = 0; j < K; ++j) {
      int val = cur->sum.a[i][j];
      if (val == INF) val = -1;
      printf("%d ", val);
    }
    puts("");
  }
  debug(cur->left, dep + 1);
  debug(cur->right, dep + 1);
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  scanf("%s", s);
  last = pool;
  Node* root = build(0, n);
  while (m--) {
    scanf("%d%d", &A, &B);
    --A;
    int ret = query(root).a[0][4];
    if (ret == INF) ret = -1;
    printf("%d\n", ret);
  }
  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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
const int INF = 1 << 29;
const int MOD = 1e9 + 7;
const int N = 2e5 + 10;
const int K = 5;
char s[N];
struct Mat {
  int a[5][5];
};
Mat operator+(const Mat& a, const Mat& b) {
  Mat ret;
  for (int i = 0; i < K; ++i) {
    for (int j = 0; j < K; ++j) {
      ret.a[i][j] = INF;
      for (int k = 0; k < K; ++k) {
        ret.a[i][j] = min(ret.a[i][j], a.a[i][k] + b.a[k][j]);
      }
    }
  }
  return ret;
}
struct Node {
  int a, b;
  Node *left, *right;
  Mat sum;
};
Node pool[N << 1], *last;
Node* build(int a, int b) {
  Node* cur = last++;
  cur->a = a;
  cur->b = b;
  if (a + 1 == b) {
    for (int i = 0; i < K; ++i) {
      for (int j = 0; j < K; ++j) {
        cur->sum.a[i][j] = INF;
      }
      cur->sum.a[i][i] = 0;
    }
    if (s[a] == '2') {
      cur->sum.a[0][0] = 1;
      cur->sum.a[0][1] = 0;
    } else if (s[a] == '0') {
      cur->sum.a[1][1] = 1;
      cur->sum.a[1][2] = 0;
    } else if (s[a] == '1') {
      cur->sum.a[2][2] = 1;
      cur->sum.a[2][3] = 0;
    } else if (s[a] == '6') {
      cur->sum.a[3][3] = 1;
      cur->sum.a[4][4] = 1;
    } else if (s[a] == '7') {
      cur->sum.a[3][3] = 1;
      cur->sum.a[3][4] = 0;
    }
    return cur;
  }
  cur->left = build(a, (a + b) / 2);
  cur->right = build((a + b) / 2, b);
  cur->sum = cur->left->sum + cur->right->sum;
  return cur;
}
int A, B;
Mat query(Node* cur) {
  if (cur->a >= A && cur->b <= B) {
    return cur->sum;
  }
  Mat ret;
  for (int i = 0; i < K; ++i) {
    for (int j = 0; j < K; ++j) {
      ret.a[i][j] = INF;
    }
    ret.a[i][i] = 0;
  }
  if ((cur->a + cur->b) / 2 > A) ret = ret + query(cur->left);
  if ((cur->a + cur->b) / 2 < B) ret = ret + query(cur->right);
  return ret;
}
void debug(Node* cur, int dep) {
  if (cur == nullptr) return;
  for (int i = 0; i < dep; ++i) putchar(' ');
  printf("[%d,%d):\n", cur->a, cur->b);
  for (int i = 0; i < K; ++i) {
    for (int j = 0; j < K; ++j) {
      int val = cur->sum.a[i][j];
      if (val == INF) val = -1;
      printf("%d ", val);
    }
    puts("");
  }
  debug(cur->left, dep + 1);
  debug(cur->right, dep + 1);
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  scanf("%s", s);
  last = pool;
  Node* root = build(0, n);
  while (m--) {
    scanf("%d%d", &A, &B);
    --A;
    int ret = query(root).a[0][4];
    if (ret == INF) ret = -1;
    printf("%d\n", ret);
  }
  return 0;
}

```