President Donald Trump will meet with the foreign leader believed to be named in a whistleblower complaint filed against the American president by an as-yet unknown U.S. intelligence official. Trump will meet with Ukrainian President Volodymyr Zelensky in New York, according to multiple sources, including AFP.

Both presidents will be in New York City next week to participate in the 74th session of the United Nations General Assembly.

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“In New York, the head of state will hold a series of bilateral meetings with foreign leaders, in particular with President of the United States Donald Trump,” Ukrinform, the Ukrainian state news agency reports.

Based on reports from multiple news outlets, especially The Washington Post, the whistleblower complaint is believed to allege Trump strong-armed Ukrainian President Zelensky to investigate and dig up dirt on his top political opponent, former Vice President Joe Biden, a Democrat, who is now running for president. The complaint also alleges Trump took multiple actions that were troubling, including making a disturbing “promise.” What that promise was is currently unknown.

President Trump’s personal attorney admitted, in a crazed rant on CNN Thursday night, that he asked Ukraine to investigate Biden and his son.

If true, experts say, Trump’s actions could be both criminal and impeachable.

The intelligence community’s inspector general characterized the whistleblower complaint as both “credible” and “urgent.”

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The Trump White House has, in direct contraction to law, experts believe, refused to give Congress the whistleblower’s complaint.