#include <bits/stdc++.h>
using namespace std;
long long n, k, l, ans;
long long a[1000001], b[1000001], c[1000001];
vector<vector<long long> > dp, dp1;
long long lft, all, nall;
long long f[1000001];
long long p(int i) { return (l / n - i + 1) % 1000000007; }
bool cmp(int x, int y) { return a[x] < a[y]; }
void go() {
  ans = n * p(1) % 1000000007;
  for (int i = 1; i <= n; i++) dp[1][i] = 1;
  for (int i = 1; i <= n; i++) dp1[1][i] = dp1[1][i - 1] + dp[1][i];
  for (int i = 2; i <= all; i++) {
    for (int j = 1; j <= n; j++) {
      dp[i][j] = (dp1[i - 1][f[j]]) % 1000000007;
      dp1[i][j] = (dp1[i][j - 1] + dp[i][j]) % 1000000007;
      ans = (ans + dp[i][j] * p(i) % 1000000007) % 1000000007;
    }
  }
  if (lft == 0) return;
  for (int i = 1; i <= all; i++)
    for (int j = 1; j <= lft; j++) ans = (ans + dp[i][f[a[j]]]) % 1000000007;
}
int main() {
  scanf("%I64d %I64d %I64d", &n, &l, &k);
  lft = l % n;
  all = min(k, l / n + (l % n != 0));
  nall = (k == all ? k : all - 1);
  dp.resize(all + 1);
  dp1.resize(all + 1);
  for (int i = 0; i <= all; i++) dp[i].resize(n + 1), dp1[i].resize(n + 1);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = i;
  sort(b + 1, b + n + 1, cmp);
  for (int i = 1; i <= n; i++) c[i] = a[b[i]];
  for (int i = 1; i <= n; i++) a[b[i]] = i;
  for (int i = 1; i <= n; i++) {
    int j = i;
    while (j <= n && c[j] == c[i]) j++;
    for (int k = i; k < j; k++) f[k] = j - 1;
    i = j - 1;
  }
  go();
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

While Duff was resting in the beach, she accidentally found a strange array b0, b1, ..., bl - 1 consisting of l positive integers. This array was strange because it was extremely long, but there was another (maybe shorter) array, a0, ..., an - 1 that b can be build from a with formula: bi = ai mod n where a mod b denoted the remainder of dividing a by b.

<image>

Duff is so curious, she wants to know the number of subsequences of b like bi1, bi2, ..., bix (0 ≤ i1 < i2 < ... < ix < l), such that: 

  * 1 ≤ x ≤ k
  * For each 1 ≤ j ≤ x - 1, <image>
  * For each 1 ≤ j ≤ x - 1, bij ≤ bij + 1. i.e this subsequence is non-decreasing. 



Since this number can be very large, she want to know it modulo 109 + 7.

Duff is not a programmer, and Malek is unavailable at the moment. So she asked for your help. Please tell her this number.

Input

The first line of input contains three integers, n, l and k (1 ≤ n, k, n × k ≤ 106 and 1 ≤ l ≤ 1018).

The second line contains n space separated integers, a0, a1, ..., an - 1 (1 ≤ ai ≤ 109 for each 0 ≤ i ≤ n - 1). 

Output

Print the answer modulo 1 000 000 007 in one line.

Examples

Input

3 5 3
5 9 1


Output

10


Input

5 10 3
1 2 3 4 5


Output

25

Note

In the first sample case, <image>. So all such sequences are: <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image> and <image>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, k, l, ans;
long long a[1000001], b[1000001], c[1000001];
vector<vector<long long> > dp, dp1;
long long lft, all, nall;
long long f[1000001];
long long p(int i) { return (l / n - i + 1) % 1000000007; }
bool cmp(int x, int y) { return a[x] < a[y]; }
void go() {
  ans = n * p(1) % 1000000007;
  for (int i = 1; i <= n; i++) dp[1][i] = 1;
  for (int i = 1; i <= n; i++) dp1[1][i] = dp1[1][i - 1] + dp[1][i];
  for (int i = 2; i <= all; i++) {
    for (int j = 1; j <= n; j++) {
      dp[i][j] = (dp1[i - 1][f[j]]) % 1000000007;
      dp1[i][j] = (dp1[i][j - 1] + dp[i][j]) % 1000000007;
      ans = (ans + dp[i][j] * p(i) % 1000000007) % 1000000007;
    }
  }
  if (lft == 0) return;
  for (int i = 1; i <= all; i++)
    for (int j = 1; j <= lft; j++) ans = (ans + dp[i][f[a[j]]]) % 1000000007;
}
int main() {
  scanf("%I64d %I64d %I64d", &n, &l, &k);
  lft = l % n;
  all = min(k, l / n + (l % n != 0));
  nall = (k == all ? k : all - 1);
  dp.resize(all + 1);
  dp1.resize(all + 1);
  for (int i = 0; i <= all; i++) dp[i].resize(n + 1), dp1[i].resize(n + 1);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = i;
  sort(b + 1, b + n + 1, cmp);
  for (int i = 1; i <= n; i++) c[i] = a[b[i]];
  for (int i = 1; i <= n; i++) a[b[i]] = i;
  for (int i = 1; i <= n; i++) {
    int j = i;
    while (j <= n && c[j] == c[i]) j++;
    for (int k = i; k < j; k++) f[k] = j - 1;
    i = j - 1;
  }
  go();
  cout << ans << endl;
  return 0;
}

```