#include <bits/stdc++.h>
using namespace std;
string str;
const int NMAX = 200000 + 5;
const int INF = 5 * NMAX;
struct Node {
  int st, dr;
  int cnt[10];
  int ans[5][5];
} tree[4 * NMAX];
void unite(Node &res, const Node a, const Node &b) {
  if (a.st == -1) {
    res = b;
    return;
  }
  for (int i = 0; i < 10; ++i) res.cnt[i] = a.cnt[i] + b.cnt[i];
  for (int i = 0; i < 5; ++i)
    for (int j = i + 1; j < 5; ++j) {
      res.ans[i][j] = INF;
      for (int k = i; k <= j; ++k)
        res.ans[i][j] = min(res.ans[i][j], a.ans[i][k] + b.ans[k][j]);
    }
  res.ans[0][0] = res.cnt[2];
  res.ans[1][1] = res.cnt[0];
  res.ans[2][2] = res.cnt[1];
  res.ans[3][3] = res.cnt[7] + res.cnt[6];
  res.ans[4][4] = res.cnt[6];
}
void build(int node, int st, int dr) {
  tree[node].st = st, tree[node].dr = dr;
  if (st == dr) {
    for (int i = 0; i < 10; ++i) tree[node].cnt[i] = 0;
    tree[node].cnt[str[st] - '0'] = 1;
    for (int i = 0; i < 5; ++i)
      for (int j = i + 1; j < 5; ++j) tree[node].ans[i][j] = INF;
    tree[node].ans[0][0] = tree[node].cnt[2];
    tree[node].ans[1][1] = tree[node].cnt[0];
    tree[node].ans[2][2] = tree[node].cnt[1];
    tree[node].ans[3][3] = tree[node].cnt[7] + tree[node].cnt[6];
    tree[node].ans[4][4] = tree[node].cnt[6];
    if (tree[node].cnt[2] == 1) tree[node].ans[0][1] = 0;
    if (tree[node].cnt[0] == 1) tree[node].ans[1][2] = 0;
    if (tree[node].cnt[1] == 1) tree[node].ans[2][3] = 0;
    if (tree[node].cnt[7] == 1) tree[node].ans[3][4] = 0;
    return;
  }
  int mid = (tree[node].st + tree[node].dr) >> 1;
  build(node << 1, st, mid);
  build((node << 1) + 1, mid + 1, dr);
  unite(tree[node], tree[node << 1], tree[(node << 1) + 1]);
}
Node ans;
void query(int node, int st, int dr) {
  if (tree[node].st == st && tree[node].dr == dr) {
    unite(ans, ans, tree[node]);
    return;
  }
  int mid = (tree[node].st + tree[node].dr) >> 1;
  if (dr <= mid)
    query(node << 1, st, dr);
  else if (st > mid)
    query((node << 1) + 1, st, dr);
  else {
    query(node << 1, st, mid);
    query((node << 1) + 1, mid + 1, dr);
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  int n, q;
  cin >> n >> q;
  cin >> str;
  str = " " + str;
  build(1, 1, n);
  while (q--) {
    int a, b;
    cin >> a >> b;
    ans.st = -1;
    query(1, a, b);
    if (ans.ans[0][4] == INF)
      cout << "-1\n";
    else
      cout << ans.ans[0][4] << '\n';
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
string str;
const int NMAX = 200000 + 5;
const int INF = 5 * NMAX;
struct Node {
  int st, dr;
  int cnt[10];
  int ans[5][5];
} tree[4 * NMAX];
void unite(Node &res, const Node a, const Node &b) {
  if (a.st == -1) {
    res = b;
    return;
  }
  for (int i = 0; i < 10; ++i) res.cnt[i] = a.cnt[i] + b.cnt[i];
  for (int i = 0; i < 5; ++i)
    for (int j = i + 1; j < 5; ++j) {
      res.ans[i][j] = INF;
      for (int k = i; k <= j; ++k)
        res.ans[i][j] = min(res.ans[i][j], a.ans[i][k] + b.ans[k][j]);
    }
  res.ans[0][0] = res.cnt[2];
  res.ans[1][1] = res.cnt[0];
  res.ans[2][2] = res.cnt[1];
  res.ans[3][3] = res.cnt[7] + res.cnt[6];
  res.ans[4][4] = res.cnt[6];
}
void build(int node, int st, int dr) {
  tree[node].st = st, tree[node].dr = dr;
  if (st == dr) {
    for (int i = 0; i < 10; ++i) tree[node].cnt[i] = 0;
    tree[node].cnt[str[st] - '0'] = 1;
    for (int i = 0; i < 5; ++i)
      for (int j = i + 1; j < 5; ++j) tree[node].ans[i][j] = INF;
    tree[node].ans[0][0] = tree[node].cnt[2];
    tree[node].ans[1][1] = tree[node].cnt[0];
    tree[node].ans[2][2] = tree[node].cnt[1];
    tree[node].ans[3][3] = tree[node].cnt[7] + tree[node].cnt[6];
    tree[node].ans[4][4] = tree[node].cnt[6];
    if (tree[node].cnt[2] == 1) tree[node].ans[0][1] = 0;
    if (tree[node].cnt[0] == 1) tree[node].ans[1][2] = 0;
    if (tree[node].cnt[1] == 1) tree[node].ans[2][3] = 0;
    if (tree[node].cnt[7] == 1) tree[node].ans[3][4] = 0;
    return;
  }
  int mid = (tree[node].st + tree[node].dr) >> 1;
  build(node << 1, st, mid);
  build((node << 1) + 1, mid + 1, dr);
  unite(tree[node], tree[node << 1], tree[(node << 1) + 1]);
}
Node ans;
void query(int node, int st, int dr) {
  if (tree[node].st == st && tree[node].dr == dr) {
    unite(ans, ans, tree[node]);
    return;
  }
  int mid = (tree[node].st + tree[node].dr) >> 1;
  if (dr <= mid)
    query(node << 1, st, dr);
  else if (st > mid)
    query((node << 1) + 1, st, dr);
  else {
    query(node << 1, st, mid);
    query((node << 1) + 1, mid + 1, dr);
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  int n, q;
  cin >> n >> q;
  cin >> str;
  str = " " + str;
  build(1, 1, n);
  while (q--) {
    int a, b;
    cin >> a >> b;
    ans.st = -1;
    query(1, a, b);
    if (ans.ans[0][4] == INF)
      cout << "-1\n";
    else
      cout << ans.ans[0][4] << '\n';
  }
  return 0;
}

```