You are given an undirected graph with n vertices. There are no edge-simple cycles with the even length in it. In other words, there are no cycles of even length that pass each edge at most once. Let's enumerate vertices from 1 to n. 

You have to answer q queries. Each query is described by a segment of vertices [l; r], and you have to count the number of its subsegments [x; y] (l ≤ x ≤ y ≤ r), such that if we delete all vertices except the segment of vertices [x; y] (including x and y) and edges between them, the resulting graph is bipartite.

Input

The first line contains two integers n and m (1 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105) — the number of vertices and the number of edges in the graph.

The next m lines describe edges in the graph. The i-th of these lines contains two integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), denoting an edge between vertices ai and bi. It is guaranteed that this graph does not contain edge-simple cycles of even length.

The next line contains a single integer q (1 ≤ q ≤ 3·105) — the number of queries.

The next q lines contain queries. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the query parameters.

Output

Print q numbers, each in new line: the i-th of them should be the number of subsegments [x; y] (li ≤ x ≤ y ≤ ri), such that the graph that only includes vertices from segment [x; y] and edges between them is bipartite.

Examples

Input

6 6
1 2
2 3
3 1
4 5
5 6
6 4
3
1 3
4 6
1 6


Output

5
5
14


Input

8 9
1 2
2 3
3 1
4 5
5 6
6 7
7 8
8 4
7 2
3
1 8
1 4
3 8


Output

27
8
19

Note

The first example is shown on the picture below:

<image>

For the first query, all subsegments of [1; 3], except this segment itself, are suitable.

For the first query, all subsegments of [4; 6], except this segment itself, are suitable.

For the third query, all subsegments of [1; 6] are suitable, except [1; 3], [1; 4], [1; 5], [1; 6], [2; 6], [3; 6], [4; 6].

The second example is shown on the picture below:

<image>
#include <bits/stdc++.h>
using namespace std;
int n, m, q, numCycles, cycle[300000], parent[300000], depth[300000],
    farthestRight[300000], firstAbove[300000];
long long preSum[300000];
pair<int, int> cycleMinMax[300000 / 3];
vector<int> connections[300000];
void dfs(int node) {
  for (const int next : connections[node]) {
    if (next == parent[node]) continue;
    if (depth[next] != -1 && depth[next] < depth[node]) {
      int curr = node;
      cycleMinMax[numCycles] = make_pair(curr, curr);
      while (1) {
        cycle[curr] = numCycles;
        cycleMinMax[numCycles].first = min(cycleMinMax[numCycles].first, curr);
        cycleMinMax[numCycles].second =
            max(cycleMinMax[numCycles].second, curr);
        if (curr == next) break;
        curr = parent[curr];
      }
      numCycles++;
    } else if (depth[next] == -1) {
      depth[next] = 1 + depth[node];
      parent[next] = node;
      dfs(next);
    }
  }
  if (cycle[node] == -1) {
    cycleMinMax[numCycles] = make_pair(node, n);
    cycle[node] = numCycles++;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < (m); i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    connections[a].push_back(b);
    connections[b].push_back(a);
  }
  for (int i = 0; i < (n); i++) cycle[i] = -1, depth[i] = -1;
  for (int i = 0; i < (n); i++)
    if (depth[i] == -1) {
      depth[i] = 0;
      parent[i] = -1;
      dfs(i);
    }
  int lptr = 0, rptr = 0;
  int curr = 0;
  while (lptr < n) {
    if (rptr == n || rptr == cycleMinMax[cycle[rptr]].second &&
                         cycleMinMax[cycle[rptr]].first >= lptr) {
      farthestRight[lptr] = rptr - 1;
      while (rptr - 1 > curr) firstAbove[curr++] = lptr;
      lptr++;
    } else
      rptr++;
  }
  firstAbove[curr] = n;
  for (int i = 0; i < (n); i++) {
    preSum[i] = farthestRight[i];
    if (i > 0) preSum[i] += preSum[i - 1];
  }
  cin >> q;
  for (int i = 0; i < (q); i++) {
    long long l, r;
    cin >> l >> r;
    l--, r--;
    long long res = 0;
    long long j = firstAbove[r] - 1;
    if (j >= l) {
      res += preSum[j];
      if (l > 0) res -= preSum[l - 1];
      res -= (j + l - 2) * (j - l + 1) / 2;
    }
    j = max(j, l - 1);
    res += (r - j + 1) * (r - j) / 2;
    cout << res << "\n";
  }
  return 0;
}
