Let $N$ be the number that consists of 61 consecutive 3's, so $N = \underbrace{333\ldots333}_{61 \, 3's}$. Let $M$ be the number that consists of 62 consecutive 6's, so $M=\underbrace{6666\ldots666}_{62 \, 6's}$. What is the digit sum of $N\times M$?

**Details and assumptions**

The **digit sum** of a number is the sum of all its digits. For example the digit sum of 1123 is $1 + 1 + 2 + 3 = 7$.

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