#include <bits/stdc++.h>
const int MAX_N = (int)2e5 + 123;
const int inf = (int)1e9 + 123;
using namespace std;
int n, q;
string second;
struct matrix {
  int dp[5][5];
  matrix() {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) dp[i][j] = (i == j ? 0 : inf);
  }
  matrix operator+(const matrix &b) {
    matrix res;
    for (int i = 0; i < 5; i++) res.dp[i][i] = inf;
    for (int i = 0; i < 5; i++)
      for (int j = i; j < 5; j++)
        for (int k = i; k <= j; k++)
          res.dp[i][j] = min(res.dp[i][j], dp[i][k] + b.dp[k][j]);
    return res;
  }
} t[4 * MAX_N];
char good[4] = {'2', '0', '1', '7'};
void build(int v = 1, int tl = 0, int tr = n - 1) {
  if (tl == tr) {
    if (second[tl] == '6') t[v].dp[3][3] = t[v].dp[4][4] = 1;
    for (int i = 0; i < 4; i++)
      if (good[i] == second[tl]) t[v].dp[i][i + 1] = 0, t[v].dp[i][i] = 1;
    return;
  }
  int tm = (tl + tr) / 2;
  build(v * 2, tl, tm), build(v * 2 + 1, tm + 1, tr);
  t[v] = t[v * 2] + t[v * 2 + 1];
}
matrix get(int l, int r, int v = 1, int tl = 0, int tr = n - 1) {
  if (tl > r || l > tr) return matrix();
  if (tl >= l && tr <= r) return t[v];
  int tm = (tl + tr) / 2;
  return get(l, r, v * 2, tl, tm) + get(l, r, v * 2 + 1, tm + 1, tr);
}
int main() {
  ios_base ::sync_with_stdio(NULL), cin.tie(NULL), cout.tie(NULL);
  cin >> n >> q >> second;
  build();
  for (int i = 1, l, r; i <= q; i++) {
    cin >> l >> r;
    matrix ans = get(l - 1, r - 1);
    if (ans.dp[0][4] == inf)
      cout << "-1\n";
    else
      cout << ans.dp[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>
const int MAX_N = (int)2e5 + 123;
const int inf = (int)1e9 + 123;
using namespace std;
int n, q;
string second;
struct matrix {
  int dp[5][5];
  matrix() {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) dp[i][j] = (i == j ? 0 : inf);
  }
  matrix operator+(const matrix &b) {
    matrix res;
    for (int i = 0; i < 5; i++) res.dp[i][i] = inf;
    for (int i = 0; i < 5; i++)
      for (int j = i; j < 5; j++)
        for (int k = i; k <= j; k++)
          res.dp[i][j] = min(res.dp[i][j], dp[i][k] + b.dp[k][j]);
    return res;
  }
} t[4 * MAX_N];
char good[4] = {'2', '0', '1', '7'};
void build(int v = 1, int tl = 0, int tr = n - 1) {
  if (tl == tr) {
    if (second[tl] == '6') t[v].dp[3][3] = t[v].dp[4][4] = 1;
    for (int i = 0; i < 4; i++)
      if (good[i] == second[tl]) t[v].dp[i][i + 1] = 0, t[v].dp[i][i] = 1;
    return;
  }
  int tm = (tl + tr) / 2;
  build(v * 2, tl, tm), build(v * 2 + 1, tm + 1, tr);
  t[v] = t[v * 2] + t[v * 2 + 1];
}
matrix get(int l, int r, int v = 1, int tl = 0, int tr = n - 1) {
  if (tl > r || l > tr) return matrix();
  if (tl >= l && tr <= r) return t[v];
  int tm = (tl + tr) / 2;
  return get(l, r, v * 2, tl, tm) + get(l, r, v * 2 + 1, tm + 1, tr);
}
int main() {
  ios_base ::sync_with_stdio(NULL), cin.tie(NULL), cout.tie(NULL);
  cin >> n >> q >> second;
  build();
  for (int i = 1, l, r; i <= q; i++) {
    cin >> l >> r;
    matrix ans = get(l - 1, r - 1);
    if (ans.dp[0][4] == inf)
      cout << "-1\n";
    else
      cout << ans.dp[0][4] << '\n';
  }
  return 0;
}

```