<p>You are given an array of <strong>positive</strong> integers <code>nums</code> of length <code>n</code>.</p>

<p>We call a pair of <strong>non-negative</strong> integer arrays <code>(arr1, arr2)</code> <strong>monotonic</strong> if:</p>

<ul>
	<li>The lengths of both arrays are <code>n</code>.</li>
	<li><code>arr1</code> is monotonically <strong>non-decreasing</strong>, in other words, <code>arr1[0] &lt;= arr1[1] &lt;= ... &lt;= arr1[n - 1]</code>.</li>
	<li><code>arr2</code> is monotonically <strong>non-increasing</strong>, in other words, <code>arr2[0] &gt;= arr2[1] &gt;= ... &gt;= arr2[n - 1]</code>.</li>
	<li><code>arr1[i] + arr2[i] == nums[i]</code> for all <code>0 &lt;= i &lt;= n - 1</code>.</li>
</ul>

<p>Return the count of <strong>monotonic</strong> pairs.</p>

<p>Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [2,3,2]</span></p>

<p><strong>Output:</strong> <span class="example-io">4</span></p>

<p><strong>Explanation:</strong></p>

<p>The good pairs are:</p>

<ol>
	<li><code>([0, 1, 1], [2, 2, 1])</code></li>
	<li><code>([0, 1, 2], [2, 2, 0])</code></li>
	<li><code>([0, 2, 2], [2, 1, 0])</code></li>
	<li><code>([1, 2, 2], [1, 1, 0])</code></li>
</ol>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [5,5,5,5]</span></p>

<p><strong>Output:</strong> <span class="example-io">126</span></p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n == nums.length &lt;= 2000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
</ul>

M=1000000007
class Solution:
    def countOfPairs(self, nums: List[int]) -> int:
        n=len(nums)
        @cache
        def solve(a,i):
            if i==len(nums):
                return 1
            temp=0
            for j in range(nums[i]+1):
                if j>=a and (i==0 or nums[i]-j<=nums[i-1]-a):
                    temp+=solve(j,i+1)
            return temp%M
        mx=max(nums)+1
        dp=[[0]*mx for i in range(n)]
        for i in range(nums[0]+1):
            dp[0][i]=1
        pre=[[0]*(mx) for i in range(n)]
        pre[0][0]=dp[0][0]

        for num in range(1,mx):
            pre[0][num]=pre[0][num-1]+dp[0][num]

        for i in range(1,n):
            temp=0
            for j in range(nums[i]+1):
                b= nums[i]-j
                if b<0:break

                ub=min(nums[i-1]-b,j)
                if ub>=0:
                    dp[i][j]= (pre[i-1][ub]+dp[i][j])%M
            pre[i][0]=dp[i][0]
            for j in range(1,mx):
                pre[i][j]=(pre[i][j-1]+dp[i][j])%M


        return sum(dp[-1])%M

152937