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>
clock_t t = clock();
namespace my_std {
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
inline T rnd(T l, T r) {
  return uniform_int_distribution<T>(l, r)(rng);
}
template <typename T>
inline bool chkmax(T& x, T y) {
  return x < y ? x = y, 1 : 0;
}
template <typename T>
inline bool chkmin(T& x, T y) {
  return x > y ? x = y, 1 : 0;
}
template <typename T>
inline void read(T& t) {
  t = 0;
  char f = 0, ch = getchar();
  double d = 0.1;
  while (ch > '9' || ch < '0') f |= (ch == '-'), ch = getchar();
  while (ch <= '9' && ch >= '0') t = t * 10 + ch - 48, ch = getchar();
  if (ch == '.') {
    ch = getchar();
    while (ch <= '9' && ch >= '0') t += d * (ch ^ 48), d *= 0.1, ch = getchar();
  }
  t = (f ? -t : t);
}
template <typename T, typename... Args>
inline void read(T& t, Args&... args) {
  read(t);
  read(args...);
}
char __sr[1 << 21], __z[20];
int __C = -1, __zz = 0;
inline void Ot() { fwrite(__sr, 1, __C + 1, stdout), __C = -1; }
inline void print(register int x) {
  if (__C > 1 << 20) Ot();
  if (x < 0) __sr[++__C] = '-', x = -x;
  while (__z[++__zz] = x % 10 + 48, x /= 10)
    ;
  while (__sr[++__C] = __z[__zz], --__zz)
    ;
  __sr[++__C] = '\n';
}
void file() {}
inline void chktime() {}
long long ksm(long long x, int y) {
  long long ret = 1;
  for (; y; y >>= 1, x = x * x)
    if (y & 1) ret = ret * x;
  return ret;
}
}  // namespace my_std
using namespace my_std;
int n, Q;
int a[4666];
int B;
vector<int> p[4666];
vector<vector<int>> id[4666];
int cnt;
pair<int, int> ans[2220202];
int merge(int x, int y) {
  if (!x || !y) return x + y;
  ++cnt;
  ans[cnt] = make_pair(x, y);
  return cnt;
}
vector<vector<int>> work(vector<int> p) {
  int n = p.size();
  vector<vector<int>> res;
  res.resize(n);
  for (int i = (0); i <= (n - 1); i++) res[i].resize(n);
  if (n == 1) return res[0][0] = p[0], res;
  vector<int> L, R;
  int l = 1e9, r = 0;
  for (auto x : p) chkmin(l, a[x]), chkmax(r, a[x]);
  int mid = (l + r) >> 1;
  for (auto x : p)
    if (a[x] <= mid)
      L.push_back(x);
    else
      R.push_back(x);
  vector<vector<int>> LL = work(L), RR = work(R);
  int n1 = L.size(), n2 = R.size();
  int l1 = 0, l2 = 0;
  for (int l = (0); l <= (n - 1); l++) {
    int r1 = l1 - 1, r2 = l2 - 1;
    for (int r = (l); r <= (n - 1); r++)
      (a[p[r]] <= mid) ? ++r1 : ++r2,
          res[l][r] = merge(((l1 <= r1 && r1 < n1) ? LL[l1][r1] : 0),
                            ((l2 <= r2 && r2 < n2) ? RR[l2][r2] : 0));
    (a[p[l]] <= mid) ? ++l1 : ++l2;
  }
  return res;
}
int ans2[4666 << 4];
int main() {
  file();
  read(n, Q);
  B = sqrt(Q);
  chkmin(B, n);
  cnt = n;
  for (int i = (1); i <= (n); i++) read(a[i]), p[a[i] / B].push_back(i);
  for (int i = (1 / B); i <= (n / B); i++) id[i] = work(p[i]);
  for (int i = (1); i <= (Q); i++) {
    int l, r;
    read(l, r);
    for (int k = (1 / B); k <= (n / B); k++) {
      int L = lower_bound(p[k].begin(), p[k].end(), l) - p[k].begin(),
          R = upper_bound(p[k].begin(), p[k].end(), r) - p[k].begin() - 1;
      if (L <= R) ans2[i] = merge(ans2[i], id[k][L][R]);
    }
  }
  printf("%d\n", cnt);
  for (int i = (n + 1); i <= (cnt); i++)
    printf("%d %d\n", ans[i].first, ans[i].second);
  for (int i = (1); i <= (Q); i++) printf("%d ", ans2[i]);
  return 0;
}