Problem: You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.

We define two kinds of functions on sets:

f(S)=min_{u∈ S} u;

g(S)=max_{u∈ S} u.

You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).

Formally, you can perform the following sequence of operations:

  * cnt← cnt+1;

  * S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).




You are required to obtain some specific sets.

There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.

In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.

Input

The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.

The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.

i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.

Output

It is guaranteed that a solution under given constraints exists.

The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.

cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.

The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.

Please notice the large amount of output.

Examples

Input


3 2
1 3 2
2 3
1 3


Output


6
3 2
1 3
5 2
4 6 

Input


2 4
2 1
1 2
1 2
1 2
1 1


Output


5
2 1
2 1
2 1
5 3 3 1

Note

In the first sample:

We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.

In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.

In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.

In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.

For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.

For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6

Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
const int MAXN = (1 << 18);
const int INF = 1e9;
const int MAXM = (1 << 22);
int N, Q;
int a[MAXN + 1];
int cnt;
pair<int, int> E[MAXM + 1];
int Ans[MAXN + 1];
vector<int> vec[MAXN << 2];
int pos[MAXN + 1];
unordered_map<long long, int> vis[MAXN << 2];
inline void build(int rt, int l, int r) {
  if (l == r) {
    vec[rt].push_back(pos[l]);
    return;
  }
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  for (int i = l; i <= r; i++) vec[rt].push_back(pos[i]);
  sort(vec[rt].begin(), vec[rt].end());
}
inline int merge(int ls, int rs) {
  if (!ls) return rs;
  if (!rs) return ls;
  E[++cnt] = make_pair(ls, rs);
  return cnt;
}
inline int Find(int rt, int l, int r, int L, int R) {
  int ls = lower_bound(vec[rt].begin(), vec[rt].end(), L) - vec[rt].begin();
  int rs = upper_bound(vec[rt].begin(), vec[rt].end(), R) - vec[rt].begin();
  --rs;
  if (ls > rs) return 0;
  if (l == r) return pos[l];
  int &x = vis[rt][(long long)ls * (N + 1) + rs];
  if (x) return x;
  int mid = (l + r) >> 1;
  int tl = Find(rt << 1, l, mid, L, R),
      tr = Find(rt << 1 | 1, mid + 1, r, L, R);
  x = merge(tl, tr);
  return x;
}
int main() {
  N = read(), Q = read();
  for (int i = 1; i <= N; i++) a[i] = read(), pos[a[i]] = i;
  build(1, 1, N);
  cnt = N;
  for (int i = 1; i <= Q; i++) {
    int l = read(), r = read();
    Ans[i] = Find(1, 1, N, l, r);
  }
  printf("%d\n", cnt);
  for (int i = N + 1; i <= cnt; i++) printf("%d %d\n", E[i].first, E[i].second);
  for (int i = 1; i <= Q; i++) printf("%d ", Ans[i]);
  return 0;
}