#include <stdio.h>
#include <string.h>
#define N 3000
#define C 3000
int max(int a, int b) { return a > b ? a : b; }
int pp[N][C], dp[N][N];
int count(int l, int r, int a, int b) {
return l > r || a > b ? 0 : pp[r][b] - (a == 0 ? 0 : pp[r][a - 1]) - (l == 0 ? 0 : pp[l - 1][b] - (a == 0 ? 0 : pp[l - 1][a - 1]));
}
int solve(int *aa, int n, int c) {
int i, i_, j_, j, j1, j2, ij, a, k, ans;
for (i = 0; i < n; i++)
memset(pp[i], 0, c * sizeof *pp[i]);
for (i = 0; i < n; i++)
pp[i][aa[i]]++;
for (i = 0; i < n; i++)
for (a = 1; a < c; a++)
pp[i][a] += pp[i][a - 1];
for (a = 0; a < c; a++)
for (i = 1; i < n; i++)
pp[i][a] += pp[i - 1][a];
for (i = 0; i < n; i++)
for (j = i; j < n; j++)
dp[i][j] = aa[i] != aa[j] ? 0 : (i > 0 && j + 1 < n ? dp[i - 1][j + 1] : 0) + 1;
ans = 0;
for (a = 0; a < c; a++)
ans = max(ans, count(0, n - 1, a, a));
for (ij = 0; ij <= (n - 1) + (n - 1); ij++) {
i = ij / 2, j = ij - i;
while (i >= 0 && j < n && aa[i] == aa[j])
i--, j++;
ans = max(ans, j - i - 1);
for (i_ = i, j_ = j, j1 = j, j2 = n - 1; i_ >= 0 && j_ < n; i_--, j_++) {
if (i_ < i && aa[i_] < aa[i_ + 1])
break;
k = count(i_, i, aa[i_], aa[i_]);
while (j1 < n && count(j, j1, aa[i_], aa[i_]) < k)
j1++;
k = count(i_, i, 0, aa[i_] - 1);
while (j2 >= j && count(j, j2, 0, aa[i_] - 1) > k)
j2--;
if (j1 <= j2) {
ans = max(ans, j_ - i_ + 1);
if (j1 <= j_ && j_ <= j2)
ans = max(ans, j_ - i_ + 1 + (i_ > 0 && j_ + 1 < n ? dp[i_ - 1][j_ + 1] * 2 : 0));
}
}
}
for (i = 0; i + 1 < n; i++) {
a = -1;
for (j = i + 1; j < n; j++)
if (aa[j] < aa[i]) {
a = aa[j];
break;
}
for (i_ = i, j1 = i + 1, j2 = n - 1; i_ >= 0; i_--) {
if (i_ < i && aa[i_] < aa[i_ + 1])
break;
k = count(i_, i, aa[i_], aa[i_]);
while (j1 < n && count(i + 1, j1, aa[i_], aa[i_]) < k)
j1++;
k = count(i_, i, 0, aa[i_] - 1);
while (j2 > i && count(i + 1, j2, 0, aa[i_] - 1) - (a == -1 ? 0 : count(i + 1, j2, a, a)) > k)
j2--;
if (j1 <= j2)
ans = max(ans, (i - i_ + 1) * 2 + (a == -1 ? 0 : count(i + 1, j2, a, a)));
}
for (i_ = i, j = i + 1; j < n; j++) {
if (aa[j] < aa[i]) {
if (aa[j] != a)
break;
} else {
k = count(i + 1, j, aa[j], aa[j]);
while (i_ >= 0 && count(i_, i, aa[j], aa[j]) < k) {
i_--;
if (i_ < 0 || aa[i_] < aa[i_ + 1]) {
i_ = -1;
break;
}
}
}
if (i_ >= 0 && i - i_ + 1 == j - i - (a == -1 ? 0 : count(i + 1, j, a, a)))
ans = max(ans, j - i_ + 1 + (i_ > 0 && j + 1 < n ? dp[i_ - 1][j + 1] * 2 : 0));
}
}
return ans;
}
int main() {
static int aa[N];
int n, c, i, j, r, tmp, ans;
scanf("%d%d", &n, &c);
for (i = 0; i < n; i++)
scanf("%d", &aa[i]), aa[i]--;
ans = 0;
for (r = 0; r < 2; r++) {
ans = max(ans, solve(aa, n, c));
for (i = 0, j = n - 1; i < j; i++, j--)
tmp = aa[i], aa[i] = aa[j], aa[j] = tmp;
for (i = 0; i < n; i++)
aa[i] = c - 1 - aa[i];
}
printf("%d\n", ans);
return 0;
}
