Problem: You are given n intervals in form [l; r] on a number line.

You are also given m queries in form [x; y]. What is the minimal number of intervals you have to take so that every point (not necessarily integer) from x to y is covered by at least one of them? 

If you can't choose intervals so that every point from x to y is covered, then print -1 for that query.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of intervals and the number of queries, respectively.

Each of the next n lines contains two integer numbers l_i and r_i (0 ≤ l_i < r_i ≤ 5 ⋅ 10^5) — the given intervals.

Each of the next m lines contains two integer numbers x_i and y_i (0 ≤ x_i < y_i ≤ 5 ⋅ 10^5) — the queries.

Output

Print m integer numbers. The i-th number should be the answer to the i-th query: either the minimal number of intervals you have to take so that every point (not necessarily integer) from x_i to y_i is covered by at least one of them or -1 if you can't choose intervals so that every point from x_i to y_i is covered.

Examples

Input


2 3
1 3
2 4
1 3
1 4
3 4


Output


1
2
1


Input


3 4
1 3
1 3
4 5
1 2
1 3
1 4
1 5


Output


1
1
-1
-1

Note

In the first example there are three queries:

  1. query [1; 3] can be covered by interval [1; 3]; 
  2. query [1; 4] can be covered by intervals [1; 3] and [2; 4]. There is no way to cover [1; 4] by a single interval; 
  3. query [3; 4] can be covered by interval [2; 4]. It doesn't matter that the other points are covered besides the given query. 



In the second example there are four queries:

  1. query [1; 2] can be covered by interval [1; 3]. Note that you can choose any of the two given intervals [1; 3]; 
  2. query [1; 3] can be covered by interval [1; 3]; 
  3. query [1; 4] can't be covered by any set of intervals; 
  4. query [1; 5] can't be covered by any set of intervals. Note that intervals [1; 3] and [4; 5] together don't cover [1; 5] because even non-integer points should be covered. Here 3.5, for example, isn't covered. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 5, LOGN = 18, maxn = 2e5 + 5;
pair<int, int> a[maxn];
int nxt[N], n, m, up[LOGN + 5][N];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d%d", &a[i].first, &a[i].second);
  sort(a, a + n);
  int idx = 0, p = -1;
  for (int i = 0; i < N; i++) {
    while (idx < n && a[idx].first == i) {
      p = max(p, a[idx].second);
      idx++;
    }
    nxt[i] = (p > i ? p : -1);
  }
  for (int i = 0; i < N; i++) up[0][i] = nxt[i];
  for (int k = 1; k <= LOGN; k++) {
    for (int i = 0; i < N; i++) {
      if (up[k - 1][i] == -1)
        up[k][i] = -1;
      else
        up[k][i] = up[k - 1][up[k - 1][i]];
    }
  }
  while (m--) {
    int qx, qy, ans = 0;
    scanf("%d%d", &qx, &qy);
    for (int k = LOGN; k >= 0; k--) {
      if (up[k][qx] == -1 || up[k][qx] > qy) continue;
      qx = up[k][qx];
      ans += (1 << k);
    }
    if (qx == qy)
      printf("%d\n", ans);
    else if (nxt[qx] >= qy)
      printf("%d\n", ans + 1);
    else
      puts("-1");
  }
}