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;
constexpr int N = 3e5 + 10;
long long n, m, u, v, p, have[N], g = 1, q[N], tree[4 * N], lazy[4 * N], ans[N],
                                  qu, beg[N];
vector<pair<int, int>> node[N], zz[N];
vector<int> adj[N], badj[N], vec;
bool mark[N], tof[N];
void dfs(int v) {
  mark[v] = true;
  for (auto i : node[v]) {
    if (!tof[i.second]) {
      adj[v].push_back(i.first);
      badj[i.first].push_back(v);
      tof[i.second] = true;
    }
    if (!mark[i.first]) dfs(i.first);
  }
}
void dfs2(int v) {
  mark[v] = true;
  for (auto i : adj[v])
    if (!mark[i]) dfs2(i);
  vec.push_back(v);
}
void dfs3(int v) {
  mark[v] = true;
  q[p]++;
  have[v] = p;
  for (auto i : badj[v]) {
    if (!mark[i]) dfs3(i);
  }
}
void shift(int b, int e, int idx) {
  int mid = (b + e) >> 1;
  int lidx = idx << 1, ridx = lidx | 1;
  lazy[lidx] += lazy[idx];
  lazy[ridx] += lazy[idx];
  tree[lidx] += (mid - b) * lazy[idx];
  tree[ridx] += (e - mid) * lazy[idx];
  lazy[idx] = 0;
}
void update(int b, int e, int l, int r, int idx) {
  if (b >= r || e <= l) return;
  if (b >= l && e <= r) {
    tree[idx] += (e - b);
    lazy[idx]++;
    return;
  }
  int mid = (b + e) >> 1, lidx = idx << 1, ridx = lidx | 1;
  shift(b, e, idx);
  update(b, mid, l, r, lidx);
  update(mid, e, l, r, ridx);
  tree[idx] = tree[lidx] + tree[ridx];
}
long long query(int b, int e, int l, int r, int idx) {
  if (b >= r || e <= l) return 0;
  if (b >= l && e <= r) return tree[idx];
  int mid = (b + e) >> 1, lidx = idx << 1, ridx = lidx | 1;
  shift(b, e, idx);
  return query(b, mid, l, r, lidx) + query(mid, e, l, r, ridx);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    node[u].push_back({v, i + 1});
    node[v].push_back({u, i + 1});
  }
  memset(have, -1, sizeof(have));
  for (int i = 1; i <= n; i++)
    if (!mark[i]) dfs(i);
  memset(mark, 0, sizeof(mark));
  for (int i = 1; i <= n; i++) {
    if (!mark[i]) {
      dfs2(i);
    }
  }
  memset(mark, 0, sizeof(mark));
  p = 0;
  for (int i = (int)vec.size() - 1; ~i; i--) {
    if (!mark[vec[i]]) {
      dfs3(vec[i]);
      p++;
    }
  }
  cin >> qu;
  for (int i = 0; i < qu; i++) {
    cin >> u >> v;
    zz[v].push_back({i, u});
  }
  for (int i = 1; i <= n; i++) {
    if (have[i] != -1) {
      if (!beg[have[i]]) beg[have[i]] = i;
      q[have[i]]--;
      if (!q[have[i]] && beg[have[i]] != i) g = max(g, beg[have[i]] + 1);
    }
    update(1, n + 1, g, i + 1, 1);
    while ((int)zz[i].size()) {
      ans[(zz[i].back()).first] =
          query(1, n + 1, (zz[i].back()).second, i + 1, 1);
      zz[i].pop_back();
    }
  }
  for (int i = 0; i < qu; i++) cout << ans[i] << '\n';
  return 0;
}
