#include <bits/stdc++.h>
using namespace std;
long long getint() {
  long long f = 1, x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return f * x;
}
map<int, int> mp;
long long n, l, k;
int a[1123456];
int sa[1123456];
long long dp[1123456];
long long f[1123456];
long long sum[1123456];
int last[1123456];
int id(int i, int j) { return i * n + j; }
int main() {
  n = getint();
  l = getint();
  k = getint();
  for (int i = 0; i < n; i++) a[i] = sa[i] = getint();
  sort(sa, sa + n);
  for (int i = 0; i < n; i++) mp[sa[i]] = i;
  for (int j = 0; j < n; j++) dp[id(1, j)] = 1;
  for (int j = 0; j < n; j++) sum[j] = j + 1;
  for (int i = n - 1; i >= 0; i--) {
    if (sa[i] != sa[i + 1])
      last[i] = i;
    else
      last[i] = last[i + 1];
  }
  f[1] = n;
  for (int i = 2; i <= k; i++) {
    for (int j = 0; j < n; j++) dp[id(i, j)] = sum[last[j]];
    sum[0] = dp[id(i, 0)];
    for (int j = 1; j < n; j++) {
      sum[j] = sum[j - 1] + dp[id(i, j)];
      if (sum[j] >= 1000000007) sum[j] -= 1000000007;
    }
    f[i] = sum[n - 1];
  }
  long long ans = 0;
  for (int i = 1; i <= k && i * n <= l; i++) {
    ans = (ans + (l / n - i + 1) % 1000000007 * f[i]) % 1000000007;
  }
  int rem = l % n;
  for (int j1 = 0; j1 < rem; j1++) {
    for (int i = 1; i <= k && i <= l / n + 1; i++) {
      int j = mp[a[j1]];
      ans = ans + dp[id(i, j)];
      while (ans >= 1000000007) ans -= 1000000007;
    }
  }
  printf("%I64d", ans);
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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 getint() {
  long long f = 1, x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return f * x;
}
map<int, int> mp;
long long n, l, k;
int a[1123456];
int sa[1123456];
long long dp[1123456];
long long f[1123456];
long long sum[1123456];
int last[1123456];
int id(int i, int j) { return i * n + j; }
int main() {
  n = getint();
  l = getint();
  k = getint();
  for (int i = 0; i < n; i++) a[i] = sa[i] = getint();
  sort(sa, sa + n);
  for (int i = 0; i < n; i++) mp[sa[i]] = i;
  for (int j = 0; j < n; j++) dp[id(1, j)] = 1;
  for (int j = 0; j < n; j++) sum[j] = j + 1;
  for (int i = n - 1; i >= 0; i--) {
    if (sa[i] != sa[i + 1])
      last[i] = i;
    else
      last[i] = last[i + 1];
  }
  f[1] = n;
  for (int i = 2; i <= k; i++) {
    for (int j = 0; j < n; j++) dp[id(i, j)] = sum[last[j]];
    sum[0] = dp[id(i, 0)];
    for (int j = 1; j < n; j++) {
      sum[j] = sum[j - 1] + dp[id(i, j)];
      if (sum[j] >= 1000000007) sum[j] -= 1000000007;
    }
    f[i] = sum[n - 1];
  }
  long long ans = 0;
  for (int i = 1; i <= k && i * n <= l; i++) {
    ans = (ans + (l / n - i + 1) % 1000000007 * f[i]) % 1000000007;
  }
  int rem = l % n;
  for (int j1 = 0; j1 < rem; j1++) {
    for (int i = 1; i <= k && i <= l / n + 1; i++) {
      int j = mp[a[j1]];
      ans = ans + dp[id(i, j)];
      while (ans >= 1000000007) ans -= 1000000007;
    }
  }
  printf("%I64d", ans);
}

```