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 MAXN = 3e5 + 10;
int n, m, par[MAXN];
vector<int> adj[MAXN];
bool vis[MAXN];
int sz;
pair<int, int> seg[MAXN];
int tmn[MAXN];
long long fen[3][MAXN];
vector<int> vec[MAXN];
int q, tl[MAXN], tr[MAXN], sec[MAXN];
long long ans[MAXN];
bool cmp(int u, int v) { return tr[u] < tr[v]; }
void add(int x, int val, int t) {
  for (x++; x < MAXN; x += x & -x) fen[t][x] += val;
}
long long get(int v, int t) {
  long long ret = 0;
  for (; v; v -= v & -v) ret += fen[t][v];
  return ret;
}
long long get(int l, int r, int t) { return get(r, t) - get(l, t); }
int depth[MAXN];
void addCycle(int v, int p) {
  int mn = p, mx = p;
  while (v != p) {
    mn = min(mn, v);
    mx = max(mx, v);
    v = par[v];
  }
  seg[sz++] = {mn, mx};
  tmn[mn] = min(tmn[mn], mx);
}
void dfs(int v, int p = -1, int de = 0) {
  depth[v] = de;
  par[v] = p;
  vis[v] = 1;
  for (int u : adj[v])
    if (u != p) {
      if (!vis[u])
        dfs(u, v, de + 1);
      else if (depth[u] < depth[v])
        addCycle(v, u);
    }
}
int main() {
  scanf("%d %d", &n, &m);
  while (m--) {
    int a, b;
    scanf("%d %d", &a, &b), a--, b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  memset(tmn, 63, sizeof(tmn));
  for (int v = 0; v < n; v++)
    if (!vis[v]) dfs(v);
  for (int i = n - 1; ~i; i--)
    if (i + 1 < n) {
      tmn[i] = min(tmn[i], tmn[i + 1]);
      if (tmn[i] < 1e6) vec[tmn[i]].push_back(i);
    }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) scanf("%d %d", &tl[i], &tr[i]), tl[i]--;
  iota(sec, sec + q, 0);
  sort(sec, sec + q, cmp);
  for (int i = 0; i < n; i++) add(i, 1, 2), add(i, n - i, 1);
  int cur = 0;
  for (int r = 1; r <= n; r++) {
    for (int l : vec[r]) {
      add(l, -1, 2);
      add(l, -(n - l), 1);
      add(l, tmn[l] - l, 0);
    }
    while (cur < q && tr[sec[cur]] == r) {
      int id = sec[cur++];
      ans[id] = get(tl[id], tr[id], 0);
      ans[id] += get(tl[id], tr[id], 1) - get(tl[id], tr[id], 2) * (n - r);
    }
  }
  for (int i = 0; i < q; i++) printf("%lld\n", ans[i]);
  return 0;
}
