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 + 13;
const int L = 18;
pair<int, int> a[N];
int to[N];
int dp[L][N];
int n, q;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    a[i] = {l, r};
  }
  sort(a, a + n);
  int j = 0;
  pair<int, int> mx = {0, -1};
  for (int i = 0; i < N; i++) {
    while (j < n && a[j].first == i) {
      mx = max(mx, {a[j].second, j++});
    }
    if (mx.first <= i)
      to[i] = -1;
    else
      to[i] = mx.second;
  }
  for (int i = 0; i < n; i++) {
    dp[0][i] = to[a[i].second];
  }
  for (int i = 1; i < L; i++) {
    for (int j = 0; j < N; j++) {
      if (dp[i - 1][j] == -1)
        dp[i][j] = -1;
      else
        dp[i][j] = dp[i - 1][dp[i - 1][j]];
    }
  }
  while (q--) {
    int ans = 1;
    int l, r;
    cin >> l >> r;
    int x = to[l];
    if (x == -1) {
      cout << -1 << "\n";
      continue;
    }
    for (int i = L - 1; i >= 0; i--) {
      int k = dp[i][x];
      if (k == -1) continue;
      if (a[k].second < r) {
        ans += 1 << i;
        x = k;
      }
    }
    if (a[x].second >= r)
      cout << ans << "\n";
    else if (dp[0][x] == -1)
      cout << -1 << "\n";
    else
      cout << ans + 1 << "\n";
  }
}