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 long long MAXN = 3e5 + 7;
struct Edge {
  long long v, num;
  Edge(long long v_, long long num_) {
    v = v_;
    num = num_;
  }
  Edge() {}
};
long long n, m;
vector<Edge> g[MAXN];
pair<long long, long long> ed[MAXN];
void read() {
  cin >> n >> m;
  for (long long i = 0; i < m; ++i) {
    cin >> ed[i].first >> ed[i].second;
    --ed[i].first;
    --ed[i].second;
    g[ed[i].first].push_back(Edge(ed[i].second, i));
    g[ed[i].second].push_back(Edge(ed[i].first, i));
  }
}
bool br[MAXN];
bool used[MAXN];
long long tin[MAXN], tout[MAXN], timer = 0;
long long dp[MAXN];
void dfs(long long u, long long parent) {
  used[u] = 1;
  tin[u] = dp[u] = timer++;
  for (auto e : g[u]) {
    if (e.v == parent) continue;
    if (used[e.v]) {
      dp[u] = min(dp[u], tin[e.v]);
    } else {
      dfs(e.v, u);
      dp[u] = min(dp[u], dp[e.v]);
      if (tin[u] < dp[e.v]) {
        br[e.num] = 1;
      }
    }
  }
}
void bridges() {
  memset(used, 0, sizeof used);
  for (long long i = 0; i < n; ++i) {
    if (!used[i]) dfs(i, -1);
  }
}
set<long long> ms[MAXN];
void dfs1(long long u, long long c) {
  used[u] = 1;
  ms[c].insert(u);
  for (auto e : g[u]) {
    if (br[e.num]) continue;
    if (!used[e.v]) dfs1(e.v, c);
  }
}
long long cnt = 0;
void comps() {
  memset(used, 0, sizeof used);
  for (long long i = 0; i < n; ++i) {
    if (!used[i]) dfs1(i, cnt++);
  }
}
long long r[MAXN];
long long post[MAXN];
long long pref[MAXN];
void solve() {
  bridges();
  comps();
  for (long long i = 0; i < n; ++i) r[i] = MAXN;
  for (long long i = 0; i < cnt; ++i) {
    if (ms[i].size() > 1) {
      long long nl = *ms[i].begin();
      long long nr = *ms[i].rbegin();
      r[nl] = nr;
    }
  }
  post[n] = MAXN;
  for (long long i = n - 1; i >= 0; --i) {
    post[i] = min(post[i + 1], r[i]);
  }
  for (long long i = 0; i < n; ++i) {
    pref[i + 1] = pref[i] + post[i];
  }
}
long long getsum(long long n) { return n * (n + 1) / 2; }
long long getsum(long long l, long long r) { return getsum(r) - getsum(l - 1); }
void print() {
  long long q;
  cin >> q;
  for (long long i = 0; i < q; ++i) {
    long long l, r;
    cin >> l >> r;
    --l;
    --r;
    long long t = upper_bound(post + l, post + r + 1, r) - post;
    cout << (r + 1) * (r - t + 1) + (pref[t] - pref[l]) - getsum(l, r) << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  read();
  solve();
  print();
  return 0;
}
