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 = 300000 + 10;
int head[maxn], to[maxn << 1], nxt[maxn << 1], tot;
void AddEdge(int u, int v) {
  to[++tot] = v;
  nxt[tot] = head[u];
  head[u] = tot;
}
struct Node {
  int l, r;
  bool operator<(const Node& tmp) const {
    return l == tmp.l ? r < tmp.r : l < tmp.l;
  }
};
vector<Node> vec;
int sta[maxn], top;
int dps[maxn], vis[maxn << 1];
void dfs(int u, int fath, int tdps) {
  if (dps[u]) {
    int minv = u, maxv = u;
    for (int i = dps[u]; i < tdps; i++) {
      minv = min(minv, sta[i]);
      maxv = max(maxv, sta[i]);
    }
    vec.push_back((Node){minv, maxv});
  } else {
    sta[++top] = u;
    dps[u] = dps[fath] + 1;
    for (int i = head[u]; i; i = nxt[i]) {
      int v = to[i];
      if (v != fath && vis[i] != 1) {
        vis[i] = 1;
        (i & 1 ? vis[i + 1] : vis[i - 1]) = 1;
        dfs(v, u, tdps + 1);
      }
    }
    top--;
  }
}
int max_r[maxn];
long long sum[maxn];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    AddEdge(u, v);
    AddEdge(v, u);
  }
  for (int i = 1; i <= n; i++) {
    if (!dps[i]) {
      dfs(i, 0, 1);
    }
  }
  set<int> st;
  for (int i = 0; i < vec.size(); i++) {
    st.insert(vec[i].r);
  }
  sort(vec.begin(), vec.end());
  int now = 0;
  for (int i = 1; i <= n; i++) {
    max_r[i] = st.size() ? ((*st.begin()) - 1) : n;
    while (now < vec.size() && vec[now].l <= i) {
      st.erase(vec[now].r);
      now++;
    }
  }
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + max_r[i] - i + 1;
  }
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    int pos = upper_bound(max_r + 1, max_r + 1 + n, r) - max_r - 1;
    pos = max(pos, l - 1);
    long long ans = sum[pos] - sum[l - 1] + 1LL * (r - pos + 1) * (r - pos) / 2;
    printf("%lld\n", ans);
  }
  getchar();
  getchar();
  return 0;
}
