#include <bits/stdc++.h>
using namespace std;
const int N = 3e5;
struct mt {
  int a[5][5];
  mt() {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) a[i][j] = 0;
  }
};
mt func(mt a, mt b) {
  mt c;
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) c.a[i][j] = int(1e9);
  for (int i = 0; i < 5; i++)
    for (int j = i; j < 5; j++)
      for (int k = j; k < 5; k++)
        c.a[i][k] = min(c.a[i][k], a.a[i][j] + b.a[j][k]);
  return c;
}
mt gen(char ch) {
  mt c;
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) {
      c.a[i][j] = int(1e9);
      if (i == j) c.a[i][j] = 0;
    }
  if (ch == '2') {
    c.a[0][1] = 0;
    c.a[0][0] = 1;
  } else if (ch == '0') {
    c.a[1][1] = 1;
    c.a[1][2] = 0;
  } else if (ch == '1') {
    c.a[2][2] = 1;
    c.a[2][3] = 0;
  } else if (ch == '7') {
    c.a[3][3] = 1;
    c.a[3][4] = 0;
  } else if (ch == '6') {
    c.a[3][3] = 1;
    c.a[4][4] = 1;
  }
  return c;
}
mt t[N * 4];
string str;
void build(int v, int l, int r) {
  if (l == r) {
    t[v] = gen(str[l]);
    return;
  }
  int c = (l + r) >> 1;
  build(v << 1, l, c);
  build((v << 1) ^ 1, c + 1, r);
  t[v] = func(t[v << 1], t[(v << 1) ^ 1]);
}
mt ret;
void get(int ll, int rr, int l, int r, int v) {
  if (rr < l || r < ll) return;
  if (ll <= l && r <= rr) {
    ret = func(ret, t[v]);
    return;
  }
  int c = (l + r) >> 1;
  get(ll, rr, l, c, v << 1);
  get(ll, rr, c + 1, r, (v << 1) ^ 1);
}
void print(mt ret) {
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      cout << ret.a[i][j] << " ";
    }
    cout << "\n";
  }
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  int n, q;
  cin >> n >> q;
  cin >> str;
  build(1, 0, n - 1);
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) {
        ret.a[i][j] = int(1e9);
        if (i == j) ret.a[i][j] = 0;
      }
    get(l, r, 0, n - 1, 1);
    if (ret.a[0][4] == int(1e9))
      cout << "-1\n";
    else
      cout << ret.a[0][4] << "\n";
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 3e5;
struct mt {
  int a[5][5];
  mt() {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) a[i][j] = 0;
  }
};
mt func(mt a, mt b) {
  mt c;
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) c.a[i][j] = int(1e9);
  for (int i = 0; i < 5; i++)
    for (int j = i; j < 5; j++)
      for (int k = j; k < 5; k++)
        c.a[i][k] = min(c.a[i][k], a.a[i][j] + b.a[j][k]);
  return c;
}
mt gen(char ch) {
  mt c;
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) {
      c.a[i][j] = int(1e9);
      if (i == j) c.a[i][j] = 0;
    }
  if (ch == '2') {
    c.a[0][1] = 0;
    c.a[0][0] = 1;
  } else if (ch == '0') {
    c.a[1][1] = 1;
    c.a[1][2] = 0;
  } else if (ch == '1') {
    c.a[2][2] = 1;
    c.a[2][3] = 0;
  } else if (ch == '7') {
    c.a[3][3] = 1;
    c.a[3][4] = 0;
  } else if (ch == '6') {
    c.a[3][3] = 1;
    c.a[4][4] = 1;
  }
  return c;
}
mt t[N * 4];
string str;
void build(int v, int l, int r) {
  if (l == r) {
    t[v] = gen(str[l]);
    return;
  }
  int c = (l + r) >> 1;
  build(v << 1, l, c);
  build((v << 1) ^ 1, c + 1, r);
  t[v] = func(t[v << 1], t[(v << 1) ^ 1]);
}
mt ret;
void get(int ll, int rr, int l, int r, int v) {
  if (rr < l || r < ll) return;
  if (ll <= l && r <= rr) {
    ret = func(ret, t[v]);
    return;
  }
  int c = (l + r) >> 1;
  get(ll, rr, l, c, v << 1);
  get(ll, rr, c + 1, r, (v << 1) ^ 1);
}
void print(mt ret) {
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      cout << ret.a[i][j] << " ";
    }
    cout << "\n";
  }
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  int n, q;
  cin >> n >> q;
  cin >> str;
  build(1, 0, n - 1);
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) {
        ret.a[i][j] = int(1e9);
        if (i == j) ret.a[i][j] = 0;
      }
    get(l, r, 0, n - 1, 1);
    if (ret.a[0][4] == int(1e9))
      cout << "-1\n";
    else
      cout << ret.a[0][4] << "\n";
  }
  return 0;
}

```