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;
const int N = 3e5;
long long tree[4 * N + 10], lazy[4 * N + 10], ans[N + 10];
int vis[N + 10], par[N + 10], cyc[N + 10];
vector<int> g[N + 10];
vector<pair<int, int> > Q[N + 10];
int find(int a) {
  if (par[a] != a) {
    return par[a] = find(par[a]);
  }
  return a;
}
void push(int v, int tl, int tr) {
  int tm = (tl + tr) / 2;
  tree[v * 2] += 1ll * (tm - tl + 1) * lazy[v];
  tree[v * 2 + 1] += 1ll * (tr - tm) * lazy[v];
  lazy[v * 2] += lazy[v];
  lazy[v * 2 + 1] += lazy[v];
  lazy[v] = 0;
}
void update(int v, int tl, int tr, int l, int r, int val) {
  if (l > r) {
    return;
  }
  if (tl == l && tr == r) {
    tree[v] += (1ll * (tr - tl + 1) * val);
    lazy[v] += val;
    return;
  }
  push(v, tl, tr);
  int tm = (tl + tr) / 2;
  update(v * 2, tl, tm, l, min(r, tm), val);
  update(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, val);
  tree[v] = tree[v * 2] + tree[v * 2 + 1];
}
long long query(int v, int tl, int tr, int l, int r) {
  if (l > r) {
    return 0;
  }
  if (tl == l && tr == r) {
    return tree[v];
  }
  push(v, tl, tr);
  int tm = (tl + tr) / 2;
  return query(v * 2, tl, tm, l, min(r, tm)) +
         query(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
}
long long cou(long long a) { return (a * (a + 1)) / 2; }
void dfs(int v, int p) {
  vis[v] = 1;
  for (auto it : g[v]) {
    if (it == p) {
      continue;
    }
    if (!vis[it]) {
      par[it] = v;
      dfs(it, v);
    } else if (vis[it] == 1) {
      int mx = v, mn = v, cur = v;
      while (cur != it) {
        mx = max(mx, par[cur]);
        mn = min(mn, par[cur]);
        cur = par[cur];
      }
      cyc[mx] = mn;
    }
  }
  vis[v] = 2;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    Q[r].push_back(make_pair(l, i));
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i, 0);
    }
  }
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    update(1, 1, n, 1, mx, 1);
    if (cyc[i] > mx) {
      update(1, 1, n, mx + 1, cyc[i], 1);
      mx = cyc[i];
    }
    for (auto it : Q[i]) {
      ans[it.second] = cou(i - it.first + 1);
      ans[it.second] -= query(1, 1, n, it.first, i);
    }
  }
  for (int i = 1; i <= q; i++) {
    printf("%lld\n", ans[i]);
  }
}
