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;
const int N = (1 << 12) + 5, Q = (1 << 16) + 5;
int n, q, l, r, cnt, a[N], ans[Q], inv[N];
long long rnd[N];
vector<pair<int, int>> ve;
mt19937_64 mt(133769420);
struct segment_tree {
  long long cur[4 * N];
  unordered_map<long long, int> ind[4 * N];
  void build(int l, int r, int i) {
    if (l == r) {
      ind[i][rnd[l]] = inv[l];
      return;
    }
    build(l, (l + r) / 2, i * 2);
    build((l + r) / 2 + 1, r, i * 2 + 1);
  }
  void update(int l, int r, int i, int u) {
    cur[i] ^= rnd[u];
    if (l == r) {
      return;
    }
    if (u <= (l + r) / 2) {
      update(l, (l + r) / 2, i * 2, u);
    } else {
      update((l + r) / 2 + 1, r, i * 2 + 1, u);
    }
  }
  int get_set(int i) {
    if (cur[i] == 0) {
      return 0;
    }
    if (ind[i][cur[i]] == 0) {
      int le = get_set(i * 2), ri = get_set(i * 2 + 1);
      if (min(le, ri) == 0) {
        ind[i][cur[i]] = max(le, ri);
      } else {
        ve.push_back({le, ri});
        ind[i][cur[i]] = ++cnt;
      }
    }
    return ind[i][cur[i]];
  }
} seg;
struct query {
  int l, r, ind;
  query(int _l, int _r, int _ind) : l(_l), r(_r), ind(_ind) {}
};
vector<query> que;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> q;
  cnt = n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    inv[a[i]] = i;
    rnd[i] = mt();
  }
  seg.build(1, n, 1);
  for (int i = 1; i <= q; i++) {
    cin >> l >> r;
    que.push_back(query(l, r, i));
  }
  int d = max(1.0, n / sqrt(q));
  sort(que.begin(), que.end(), [&](const query& u, const query& v) {
    return u.l / d != v.l / d ? u.l < v.l : u.r < v.r;
  });
  int l = 1, r = 0;
  for (query& cur : que) {
    for (; l > cur.l; l--) {
      seg.update(1, n, 1, a[l - 1]);
    }
    for (; l < cur.l; l++) {
      seg.update(1, n, 1, a[l]);
    }
    for (; r > cur.r; r--) {
      seg.update(1, n, 1, a[r]);
    }
    for (; r < cur.r; r++) {
      seg.update(1, n, 1, a[r + 1]);
    }
    ans[cur.ind] = seg.get_set(1);
  }
  cout << cnt << '\n';
  for (auto& [l, r] : ve) {
    cout << l << " " << r << '\n';
  }
  for (int i = 1; i <= q; i++) {
    cout << ans[i] << " ";
  }
}